标签：sign line sim har query define 交换 优化 top
题面:bzoj
题解：神奇的\(dp\)
先按长度把边排序
指定必须走前\(l\)条边，枚举\(l\)
设\(dis[i][j][k]\)表示当前到了\(i\)节点，已走过了\(j\)条前\(l\)条边，用了\(k\)次交换次数
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题面：bzoj
题解：更为神奇
首先直接bfs，到一定程度break
其次乱贪心
据\(\text{M}\)\(\color{red}{\text{_sea}}\)说正解是整数规划？
不会，告辞
code
题面：Luogu
题解：斜率优化
首先有一个显然的性质就是：一段两端必定相同并且是这一段选定的大小（否则可以证明结果必定小于这种）
设\(dp[i]\)表示前\(i\)个的答案，\(s[i]\)表示\(i\)是\(1\sim n\)内相同大小的第几个
\[dp[i]=max_{0<j\leq i}^{a[i]==a[j]}{\{dp[j-1]+a[i]*(s[i]-s[j]+1)^2\}}\]
于是有
只与\(i\)有关的：\(a[i]*s[i]^2+2*a[i]*s[i]+a[i]\)
只与\(j\)有关的：\(dp[j-1]+a[i]*s[j]^2-2*a[i]*s[j](a[i]==a[j])\)
既和\(i\)有关又和\(j\)有关的：\(-2a[i]s[i]s[j]\)
于是套斜率优化板子，维护上凸壳，栈顶最优决策
注意要先把自己丢进去再决策
由于空间问题用\(vector\)模拟栈
code
题面：Luogu
题解：容斥
设至少\(x\)个人什么都没有
也就是把\(a[i]\)个特产分给\(n-x\)个人
于是在\(n-x-1\)个空格里随意插\(a[i]\)块隔板
根据乘法原理，对每一种特产考虑最后再乘起来
方案数就是\(\prod_{i=1}^{m}{C_{n-x-1+a[i]}^{a[i]}}\)
反演一下（相比之下我更喜欢这个而不是容斥）
答案就是恰好0人什么都没有
\[ans=\sum_{i=0}^{n-1}{(-1)^iC_{n}^{i}\prod_{j=1}^{m}{C_{n-i-1+a[j]}^{a[j]}}}\]
code
题面：Luogu
是重题
建一颗主席树，如果\([1,mid]\)内没有这么多个数字就显然没有，再看\([mid+1,r]\)
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题面：bzoj
题解：可并堆
修改操作就把这个点抠出来（合并它的左右儿子），再塞回去
据说可以用pb_ds水过去（然而我搞了一下午没搞出来）
code
#include<bits/stdc++.h>
using namespace std;
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout);
inline void read(int& x)
{
x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
}
#define maxn 155
struct Edge
{
int fr,to,val;
}eg[maxn<<1];
int head[maxn],edgenum;
inline void add(int fr,int to,int val)
{
eg[++edgenum]=(Edge){head[fr],to,val};
head[fr]=edgenum;
}
struct Road
{
int l,r,val;
inline friend bool operator < (Road a,Road b)
{
return a.val<b.val;
}
}d[maxn];
struct Node
{
int x,y,z,dis;
inline friend bool operator < (Node a,Node b)
{
return a.dis>b.dis;
}
};
priority_queue<Node> q;
int dis[52][maxn][maxn],vis[52][maxn][maxn];
int n,m,k,ans=0x7fffffff;
inline void work(int l)
{
memset(vis,0,sizeof(vis));
memset(dis,0x3f,sizeof(dis));
dis[1][0][0]=0;
q.push((Node){1,0,0});
int x,y,z;
while(!q.empty())
{
x=q.top().x,y=q.top().y,z=q.top().z;
q.pop();
if(vis[x][y][z]) continue;
vis[x][y][z]=1;
for(int i=head[x];i;i=eg[i].fr)
{
#define to eg[i].to
if(i<=(l<<1))
{
if(y<l&&dis[to][y+1][z]>dis[x][y][z]+d[y+1].val)
dis[to][y+1][z]=dis[x][y][z]+d[y+1].val,q.push((Node){to,y+1,z,dis[to][y+1][z]});
}
else
{
if(y<l&&z<k&&dis[to][y+1][z+1]>dis[x][y][z]+d[y+1].val)
dis[to][y+1][z+1]=dis[x][y][z]+d[y+1].val,q.push((Node){to,y+1,z+1,dis[to][y+1][z+1]});
if(dis[to][y][z]>dis[x][y][z]+eg[i].val)
dis[to][y][z]=dis[x][y][z]+eg[i].val,q.push((Node){to,y,z,dis[to][y][z]});
}
#undef to
}
}
for(int i=0;i<=k;++i) ans=min(ans,dis[n][l][i]);
}
int main()
{
read(n),read(m),read(k);
for(int i=1;i<=m;++i) read(d[i].l),read(d[i].r),read(d[i].val);
sort(d+1,d+m+1);
for(int i=1;i<=m;++i) add(d[i].l,d[i].r,d[i].val),add(d[i].r,d[i].l,d[i].val);
for(int i=0;i<=m;++i) work(i);
printf("%d\n",ans);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pa pair<ll,ll>
#define mp make_pair
#define inf 0x7fffffff
queue<pa> q;
//unordered_map<ll,bool> m;
map<ll,bool> m;
ll n,a[105],x;
inline ll bfs()
{
q.push(mp(0,0));
ll val,step,tp;
while(!q.empty())
{
val=q.front().first,step=q.front().second;
q.pop();
for(int i=1;i<=n;++i)
{
if(q.size()>=1000000) return inf;
val<=x?tp=val+a[i]:tp=val-a[i];
if(tp==x) return step+1;
if(m[tp]) continue;
m[tp]=1;
q.push(mp(tp,step+1));
}
}
return inf;
}
inline ll greedy()
{
ll step=0,sum=x;
for(int i=n;i;--i)
{
if(!a[i]) continue;
step+=sum/a[i];
sum-=sum/a[i]*a[i];
}
return sum?inf:step;
}
int main()
{
//freopen("test.in","r",stdin);
scanf("%lld",&x);
if(!x) puts("0");
while(scanf("%lld",&a[++n])!=EOF);
sort(a+1,a+n+1);
printf("%lld\n",min(bfs(),greedy()));
return 0;
}
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline void read(int& x)
{
x = 0; char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 100005
#define maxm 10005
#define ll long long
int las[maxm];
vector<int> st[maxm];
int a[maxn], s[maxn], n;
ll dp[maxn];
inline int x(const int& i)
{
return a[i] * s[i];
}
inline int y(const int& i)
{
return dp[i - 1] + (a[i] * s[i] - 2) * s[i];
}
inline ll Pow(const int& X)
{
return 1ll * X * X;
}
inline double slope(int i, int j)
{
return 1.0 * (y(i) - y(j)) / (x(i) - x(j));
}
inline ll calc(int i, int j)
{
return dp[j - 1] + a[i] * Pow(s[i] - s[j] + 1);
}
signed main()
{
read(n);
for (int i = 1; i <= n; ++i)
read(a[i]), s[i] = s[las[a[i]]] + 1, las[a[i]] = i;
for (int i = 1; i <= n; ++i)
{
#define t a[i]
#define A st[t][st[t].size()-2]
#define B st[t][st[t].size()-1]
while (st[t].size() >= 2 && slope(A, i) >= slope(A, B)) st[t].pop_back();
st[t].push_back(i);
while (st[t].size() >= 2 && calc(i, A) >= calc(i, B)) st[t].pop_back();
dp[i] = calc(i, st[t].back());
}
printf("%lld\n", dp[n]);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
inline void read(int& x)
{
x = 0; char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define p 1000000007
#define maxn 2005
int fac[maxn], inv[maxn], f[maxn], a[maxn];
inline int qpow(int x, int y)
{
int ans = 1;
while (y)
{
if (y & 1) ans = 1ll * ans * x % p;
x = 1ll * x * x % p;
y >>= 1;
}
return ans;
}
inline int C(int n, int m)
{
return 1ll * fac[n] * inv[m] % p * inv[n - m] % p;
}
int main()
{
int ans = 0, n, m;
read(n), read(m);
for (int i = 1; i <= m; ++i) read(a[i]);
fac[0] = inv[0] = 1;
for (int i = 1; i <= 2000; ++i) fac[i] = 1ll * i * fac[i - 1] % p;
inv[2000] = qpow(fac[2000], p - 2);
for (int i = 1999; i; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % p;
for (int i = 0; i <
<span id="codeI"></span>n; ++i)
{
f[0] = 1;
for (int j = 1; j <= m; ++j) f[j] = 1ll * f[j - 1] * C(n - i - 1 + a[j], n - i - 1) % p;
if (i & 1) ans = (ans - 1ll * f[m] * C(n, i) % p + p) % p;
else ans = (ans + 1ll * f[m] * C(n, i) % p) % p;
}
printf("%d\n", ans);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout);
inline void read(int& x)
{
x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
}
#define maxn 500005
int ls[maxn*32],rs[maxn*32],val[maxn*32],rt[maxn],tot;
int n,m,w;
void build(int& rt,int l,int r)
{
rt=++tot;
if(l==r) return;
int mid=(l+r)>>1;
build(ls[rt],l,mid);
build(rs[rt],mid+1,r);
}
void update(int& rt,int las,int l,int r,int pos)
{
rt=++tot;
ls[rt]=ls[las],rs[rt]=rs[las],val[rt]=val[las]+1;
if(l==r) return;
int mid=(l+r)>>1;
if(pos<=mid) update(ls[rt],ls[las],l,mid,pos);
else update(rs[rt],rs[las],mid+1,r,pos);
}
int query(int fr,int to,int len)
{
int l=1,r=n,mid;
while(true)
{
if(l==r) return l;
mid=(l+r)>>1;
if(((val[ls[to]]-val[ls[fr]])<<1)>len) r=mid,fr=ls[fr],to=ls[to];
else if(((val[rs[to]]-val[rs[fr]])<<1)>len) l=mid+1,fr=rs[fr],to=rs[to];
else return 0;
}
}
int main()
{
read(n),read(m);
build(rt[0],1,n);
for(int i=1;i<=n;++i)
read(w),update(rt[i],rt[i-1],1,n,w);
int l,r;
for(int i=1;i<=m;++i)
read(l),read(r),printf("%d\n",query(rt[l-1],rt[r],r-l+1));
return 0;
}
#include<bits/stdc++.h>
using namespace std;
inline void read(int& x)
{
x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
}
#define maxn 300005
int dis[maxn],rt[maxn],val[maxn],fa[maxn],son[maxn][2];
int merge(int x,int y)
{
if(!x||!y) return x|y;
if(val[x]>val[y]) swap(x,y);
son[x][1]=merge(son[x][1],y);
if(dis[son[x][0]]<dis[son[x][1]]) swap(son[x][0],son[x][1]);
fa[son[x][0]]=fa[son[x][1]]=x;
dis[x]=dis[son[x][1]]+1;
return x;
}
void update(int x,int y,int v)
{
int f=fa[y],tmp=merge(son[y][0],son[y][1]);
if(rt[x]==y) rt[x]=tmp;
else if(son[f][0]==y) son[f][0]=tmp;
else son[f][1]=tmp;
fa[tmp]=f;
fa[y]=son[y][0]=son[y][1]=0;
val[y]+=v;
rt[x]=merge(rt[x],y);
}
int main()
{
//freopen("test.in","r",stdin);
int n,m,k;char op[10];
read(n),read(m),read(k);
int a,b,c;
for(int i=1;i<=k;++i)
{
scanf("%s",op);
read(a);
if(op[0]=='A')
{
read(b),read(c);
val[b]=c;
rt[a]=merge(rt[a],b);
}
else if(op[0]=='D')
{
read(b),read(c);
update(a,b,-c);
}
else if(op[0]=='T')
{
read(b);
rt[b]=merge(rt[a],rt[b]);
rt[a]=0;
}
else if(op[0]=='M')
printf("%d\n",val[rt[a]]);
else
{
read(b);
if((son[rt[a]][0]&&val[son[rt[a]][0]]==val[rt[a]])||(son[rt[a]][1]&&val[son[rt[a]][1]]==val[rt[a]])) puts("ERROR");
else update(a,rt[a],b);
}
}
return 0;
}标签：sign line sim har query define 交换 优化 top
原文地址：https://www.cnblogs.com/123789456ye/p/12250181.html
