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2018-2019 ACM-ICPC Brazil Subregional Programming Contest

时间:2018-11-30 21:46:25      阅读:186      评论:0      收藏:0      [点我收藏+]

标签:names   stdout   最大流   hash   sort   temp   amp   with   签到   

A:留坑
B:二维sg函数,特判边界情况

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100+10,maxn=200000+10,inf=0x3f3f3f3f;

int sg[N][N],Hash[200];
int main()
{
    for(int i=1;i<=100;i++)
    {
        for(int j=1;j<=100;j++)
        {
            if(i==j)continue;
            memset(Hash,0,sizeof Hash);
            for(int k=1;k<i;k++)if(i-k!=j)Hash[sg[i-k][j]]=1;
            for(int k=1;k<j;k++)if(i!=j-k)Hash[sg[i][j-k]]=1;
            for(int k=1;k<min(i,j);k++)if(i-k!=j-k)Hash[sg[i-k][j-k]]=1;
            for(int k=0;;k++)
            {
                if(!Hash[k])
                {
                    sg[i][j]=k;
                    break;
                }
            }
//            printf("%d ",sg[i][j]);
        }
//        puts("");
    }
    int n,ans=0;scanf("%d",&n);
    while(n--)
    {
        int x,y;scanf("%d%d",&x,&y);
        if(x==y)return 0*puts("Y");
        ans^=sg[x][y];
    }
    puts(ans?"Y":"N");
    return 0;
}
/********************

********************/

C:bit维护逆序对即可

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=4e5+10,maxn=1000000+10,inf=0x3f3f3f3f;

struct bit{
    int sum[N];
    void init(){memset(sum,0,sizeof sum);}
    void update(int i)
    {
        for(;i<N;i+=i&(-i))sum[i]++;
    }
    int query(int i)
    {
        int ans=0;
        for(;i;i-=i&(-i))ans+=sum[i];
        return ans;
    }
}b;
pii x[N],y[N];
int Hash[N],cnt;
int main()
{
    int xx,yy,h,v;scanf("%d%d%d%d",&xx,&yy,&h,&v);
    for(int i=1;i<=h;i++)scanf("%d%d",&x[i].fi,&x[i].se),Hash[cnt++]=x[i].fi,Hash[cnt++]=x[i].se;
    sort(x+1,x+1+h);sort(Hash,Hash+cnt);cnt=unique(Hash,Hash+cnt)-Hash;
    for(int i=1;i<=h;i++)
    {
        x[i].fi=lower_bound(Hash,Hash+cnt,x[i].fi)-Hash+1;
        x[i].se=lower_bound(Hash,Hash+cnt,x[i].se)-Hash+1;
    }
    cnt=0;
    for(int i=1;i<=v;i++)scanf("%d%d",&y[i].fi,&y[i].se),Hash[cnt++]=y[i].fi,Hash[cnt++]=y[i].se;
    sort(y+1,y+1+v);sort(Hash,Hash+cnt);cnt=unique(Hash,Hash+cnt)-Hash;
    for(int i=1;i<=v;i++)
    {
        y[i].fi=lower_bound(Hash,Hash+cnt,y[i].fi)-Hash+1;
        y[i].se=lower_bound(Hash,Hash+cnt,y[i].se)-Hash+1;
    }
    ll c=0,d=0;
    for(int i=1;i<=h;i++)
    {
        c+=b.query(N-1)-b.query(x[i].se);
        b.update(x[i].se);
    }
    b.init();
    for(int i=1;i<=v;i++)
    {
        d+=b.query(N-1)-b.query(y[i].se);
        b.update(y[i].se);
    }
    printf("%lld\n",1ll*(h+1)*(v+1)+c+d);
    return 0;
}
/********************

********************/

D:签到

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=20200+10,maxn=200000+10,inf=0x3f3f3f3f;


int main()
{
    int _,x,ans=0;
    for(scanf("%d",&_);_;_--)
    {
        scanf("%d",&x);
        if(x!=1)ans++;
    }
    printf("%d\n",ans);
    return 0;
}
/********************

********************/

E:签到

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=10000+10,maxn=200000+10,inf=0x3f3f3f3f;

char s[N],p[N];
int main()
{
    scanf("%s%s",s,p);
    int ls=strlen(s),ps=strlen(p),ans=0;
    for(int i=0;i+ps-1<ls;i++)
    {
        bool f=0;
        for(int j=0;j<ps;j++)if(s[i+j]==p[j]){f=1;break;}
        if(!f)ans++;
    }
    printf("%d\n",ans);
    return 0;
}
/********************

********************/

F:注意到n很小,开(1<<n)个bit,把区间离散化后按左端点排序,bit维护区间最小值更新即可

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=2000+10,maxn=100000+10,inf=0x3f3f3f3f;

struct bit{
    int sum[N];
    bit(){memset(sum,-1,sizeof sum);}
    void update(int i,int x)
    {
        for(;i<N;i+=i&(-i))sum[i]=max(sum[i],x);
    }
    int query(int i)
    {
        int ma=-1;
        for(;i;i-=i&(-i))ma=max(ma,sum[i]);
        return ma;
    }
}b[N];
struct node{
    int s,t,v,id;
    bool operator <(const node&rhs)const
    {
        return s<rhs.s || s==rhs.s&&t<rhs.t;
    }
}p;
vector<node>v;
int Hash[N],cnt;
int main()
{
    b[0].update(1,0);
    int n;scanf("%d",&n);
    for(int i=0,k;i<n;i++)for(scanf("%d",&k);k;k--)
    {
        scanf("%d%d%d",&p.s,&p.t,&p.v),p.id=i,v.pb(p);
        Hash[cnt++]=p.s,Hash[cnt++]=p.t;
    }
    sort(Hash,Hash+cnt);cnt=unique(Hash,Hash+cnt)-Hash;
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++)
    {
        p=v[i];
        p.s=lower_bound(Hash,Hash+cnt,p.s)-Hash+1;
        p.t=lower_bound(Hash,Hash+cnt,p.t)-Hash+1;
//        printf("%d %d %d %d %d\n",p.s,p.t,p.v,p.id,b[7].query(8));
        for(int j=0;j<(1<<n);j++)
        {
            int x=b[j].query(p.s);
            if(x!=-1)b[j|(1<<p.id)].update(p.t,x+p.v);
//            if(x!=-1&&(j|(1<<p.id))==7)printf("%d %d %d!!!!\n",p.s,p.t,x+p.v);
//            if(j==5&&p.s==7&&p.id==1)printf("%d %d %d***\n",x+p.v,b[5].query(p.s),b[7].query(p.t));
        }
    }
    printf("%d\n",b[(1<<n)-1].query(N-1));
    return 0;
}
/********************

********************/

G:二分,加最大流,二分图复杂度完全可以接受

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=2000+10,maxn=50000+10,inf=0x3f3f3f3f;

int p,r,c,d[N],E[N],ans;
struct node{
    int a,b,c;
    bool operator <(const node&rhs)const
    {
        return c<rhs.c;
    }
}f[maxn];
struct edge{
    int to,Next,c;
}e[maxn];
int s,t,cnt,head[N],cur[N],dis[N];
void init(){cnt=0;memset(head,-1,sizeof head);}
void add(int u,int v,int c)
{
    e[cnt].to=v;e[cnt].c=c;e[cnt].Next=head[u];head[u]=cnt++;
    e[cnt].to=u;e[cnt].c=0;e[cnt].Next=head[v];head[v]=cnt++;
}
bool bfs()
{
    queue<int>q;
    memset(dis,-1,sizeof dis);
    dis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=head[x];~i;i=e[i].Next)
        {
            int y=e[i].to;
            if(dis[y]==-1&&e[i].c>0)
            {
                dis[y]=dis[x]+1;
                q.push(y);
            }
        }
    }
    return dis[t]!=-1;
}
int dfs(int u,int mx)
{
    if(u==t)return mx;
    int f;
    for(int &i=cur[u];~i;i=e[i].Next)
    {
        int x=e[i].to;
        if(dis[x]==dis[u]+1&&e[i].c>0&&(f=dfs(x,min(mx,e[i].c))))
        {
            e[i].c-=f;
            e[i^1].c+=f;
            return f;
        }
    }
    return 0;
}
int maxflow()
{
    int ans=0,f;
    while(bfs())
    {
        for(int i=0;i<=t;i++)cur[i]=head[i];
        while((f=dfs(s,inf)))ans+=f;
    }
    return ans;
}
bool ok(int x)
{
    init();
    for(int i=1;i<=p;i++)add(i,t,d[i]);
    for(int i=1;i<=r;i++)add(s,p+i,E[i]);
    for(int i=1;i<=c;i++)
    {
        if(f[i].c<=x)add(f[i].b+p,f[i].a,inf);
        else break;
    }
    return maxflow()==ans;
}
int main()
{
    scanf("%d%d%d",&p,&r,&c);
    s=p+r+1,t=p+r+2;
    for(int i=1;i<=p;i++)scanf("%d",&d[i]),ans+=d[i];
    for(int i=1;i<=r;i++)scanf("%d",&E[i]);
    for(int i=1;i<=c;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        f[i]={a,b,c};
    }
    sort(f+1,f+1+c);
    int l=0,r=f[c].c+1;
    while(l<r-1)
    {
        int m=(l+r)>>1;
        if(ok(m))r=m;
        else l=m;
    }
    printf("%d\n",(r<=f[c].c)?r:-1);
    return 0;
}
/********************
3 2 5
20 10 10
25 30
1 1 3
2 1 1
2 2 4
3 1 2
3 2 5
********************/

H:留坑
I:按题意模拟即可,bitset优化

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=1000+10,maxn=200000+10,inf=0x3f3f3f3f;

bitset<1010>b[N],c;
int main()
{
    int n,m,k,x;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<k;i++)scanf("%d",&x),c.set(x);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&k);
        while(k--)scanf("%d",&x),b[i].set(x);
    }
    for(int i=1;i<=n;i++)
    {
        c^=b[i];
        if(c.count()==0)return 0*printf("%d\n",i);
    }
    for(int i=1;i<=n;i++)
    {
        c^=b[i];
        if(c.count()==0)return 0*printf("%d\n",i+n);
    }
    puts("-1");
    return 0;
}
/********************

********************/

J:斯坦纳树,spfa更新时,只更新后n-k个点,保证不会出现capitals节点连capitals节点

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100+10,maxn=200000+10,inf=0x3f3f3f3f;

int x[N],y[N],n,k;
db dp[N][(1<<10)+10];
queue<int>q;
bool vis[N];
db cal(int a,int b){return sqrt(1.0*(x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
void spfa(int st)
{
    while(!q.empty())
    {
        int p=q.front();q.pop();
        vis[p]=0;
        for(int i=k;i<n;i++)
        {
            if(i==p)continue;
            db te=cal(p,i);
            if(dp[p][st]+te<dp[i][st])
            {
                dp[i][st]=dp[p][st]+te;
                q.push(i);
            }
        }
    }
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)for(int j=0;j<(1<<k);j++)dp[i][j]=2e9;
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&x[i],&y[i]);
        if(i<k)dp[i][1<<i]=0;
    }
    for(int st=0;st<(1<<k);st++)
    {
        for(int i=0;i<n;i++)
        {
            for(int s=st;s;s=(s-1)&st)
                if(dp[i][s]+dp[i][st-s]<dp[i][st])
                    dp[i][st]=dp[i][s]+dp[i][st-s];
            if(dp[i][st]<inf)q.push(i),vis[i]=1;
        }
        spfa(st);
    }
    db ans=2e9;
    for(int i=k;i<n;i++)ans=min(ans,dp[i][(1<<k)-1]);
    printf("%.5f\n",ans);
    return 0;
}
/********************

********************/

K:可以发现xy范围很小,两圆相交要满足R-r<=d,d最大是\(sqrt(2)*50\),那么我们按r sort之后,暴力枚举满足该条件的圆即可

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=150000+10,maxn=50000+10,inf=0x3f3f3f3f;

struct circle{
    db x,y,r;
    void read(){scanf("%lf%lf%lf",&x,&y,&r);}
    bool operator <(const circle&rhs)const{
        return r<rhs.r;
    }
}c[N];
bool ok(int i,int j)
{
    db d=sqrt((c[i].x-c[j].x)*(c[i].x-c[j].x)+(c[i].y-c[j].y)*(c[i].y-c[j].y));
    return c[j].r-c[i].r<d&&d<c[i].r+c[j].r;
}
int main()
{
    int n,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)c[i].read();
    sort(c+1,c+1+n);
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(c[j].r-c[i].r>sqrt(2)*50)break;
            else if(ok(i,j))
            {
                ans+=2;
                if(ans>2*n)return 0*puts("greater");
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}
/********************

********************/

L:树链剖分裸题

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=200000+10,inf=0x3f3f3f3f;

struct edge{
    int to,Next;
}e[maxn];
int cnt,head[N];
int son[N],fa[N],top[N],sz[N],id[N];
int res,dep[N];
void add(int u,int v){e[cnt].to=v;e[cnt].Next=head[u];head[u]=cnt++;}
void init()
{
    cnt=0;
    memset(head,-1,sizeof head);
    memset(son,-1,sizeof son);
}
void dfs1(int u,int f,int de)
{
    fa[u]=f;sz[u]=1;
    dep[u]=de;
    for(int i=head[u];~i;i=e[i].Next)
    {
        int v=e[i].to;
        if(v!=f)
        {
            dfs1(v,u,de+1);
            sz[u]+=sz[v];
            if(son[u]==-1||sz[v]>sz[son[u]])son[u]=v;
        }
    }
}
void dfs2(int u,int f,int tp)
{
    top[u]=tp;id[u]=++res;
    if(son[u]!=-1)dfs2(son[u],u,tp);
    for(int i=head[u];~i;i=e[i].Next)
    {
        int v=e[i].to;
        if(v!=f&&v!=son[u])dfs2(v,u,v);
    }
}
int sum[N<<2],lazy[N<<2];
void pushdown(int l,int r,int rt)
{
    if(lazy[rt]!=0)
    {
        int m=(l+r)>>1;
        sum[rt<<1]+=(m-l+1)*lazy[rt];
        sum[rt<<1|1]+=(r-m)*lazy[rt];
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        lazy[rt]=0;
    }
}
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    lazy[rt]=sum[rt]=0;
    if(l==r)return ;
    int m=(l+r)>>1;
    build(ls);build(rs);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        sum[rt]+=(r-l+1)*c;
        lazy[rt]+=c;
        return ;
    }
    pushdown(l,r,rt);
    int m=(l+r)>>1;
    if(L<=m)update(L,R,c,ls);
    if(m<R)update(L,R,c,rs);
    pushup(rt);
}
int query1(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)return sum[rt];
    pushdown(l,r,rt);
    int m=(l+r)>>1,ans=0;
    if(L<=m)ans+=query1(L,R,ls);
    if(m<R)ans+=query1(L,R,rs);
    return ans;
}
int query(int a,int b)
{
    int f1=top[a],f2=top[b],ans=0;
    while(f1!=f2)
    {
        if(dep[f1]<dep[f2])swap(f1,f2),swap(a,b);
        ans+=query1(id[f1],id[a],1,res,1);
        a=fa[f1],f1=top[a];
    }
    if(dep[a]>dep[b])swap(a,b);
    ans+=query1(id[a],id[b],1,res,1);
    return ans;
}
void change(int a,int b,int c)
{
    int f1=top[a],f2=top[b];
    while(f1!=f2)
    {
        if(dep[f1]<dep[f2])swap(f1,f2),swap(a,b);
        update(id[f1],id[a],c,1,res,1);
        a=fa[f1],f1=top[a];
    }
    if(dep[a]>dep[b])swap(a,b);
    update(id[a],id[b],c,1,res,1);
}
int main()
{
    int n,q;
    scanf("%d%d",&n,&q);
    init();
    for(int i=1;i<n;i++)
    {
        int a,b;scanf("%d%d",&a,&b);
        add(a,b);add(b,a);
    }
    dfs1(1,-1,1);
    dfs2(1,-1,1);
    build(1,res,1);
    while(q--)
    {
        int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);
        change(a,b,1);
        printf("%d\n",query(c,d));
        change(a,b,-1);
    }
    return 0;
}
/********************
10 4
1 4
4 5
3 4
3 2
7 3
6 7
7 8
10 8
8 9
6 10 2 5
1 9 5 10
9 10 2 1
5 10 2 9
********************/

M:留坑

2018-2019 ACM-ICPC Brazil Subregional Programming Contest

标签:names   stdout   最大流   hash   sort   temp   amp   with   签到   

原文地址:https://www.cnblogs.com/acjiumeng/p/10046611.html

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