标签：不能 and += inf view ring air tin begin
377A Maze
大意: 给定棋盘, 保证初始所有白格连通, 求将$k$个白格变为黑格, 使得白格仍然连通.
$dfs$回溯时删除即可.
#include <iostream> #include <functional> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head const int N = 1e3+10; int n,m,s,px,py; int vis[N][N]; char a[N][N]; const int dx[]={0,0,-1,1}; const int dy[]={-1,1,0,0}; int main() { scanf("%d%d%d", &n, &m, &s); REP(i,1,n) cin>>a[i]+1; REP(i,1,n) REP(j,1,m) if (a[i][j]==‘.‘) px=i,py=j; function<void(int,int)> dfs = [&](int x, int y) { if (x<=0||y<=0||x>n||y>m||vis[x][y]||a[x][y]==‘#‘) return; vis[x][y] = 1; REP(k,0,3) dfs(x+dx[k],y+dy[k]); if (s) --s,a[x][y]=‘X‘; }; dfs(px,py); REP(i,1,n) puts(a[i]+1); }
377B Preparing for the Contest
大意: $m$道题, $n$个学生, 每个学生每天做一道题, 每个学生只能做难度不超过他的能力的题, 雇第$i$个学生做题需要花费$j$元, 求最短时间做完所有题的情况下的最少花费, 输出方案.
二分答案, 维护一个堆贪心
#include <iostream> #include <algorithm> #include <cstdio> #include <queue> #include <functional> #define x first #define y second using namespace std; typedef long long ll; typedef pair<int,int> pii; inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} int main() { int n=rd(),m=rd(),s=rd(); vector<pii> f(m); int id = 0; for(auto &t:f) t.x=rd(),t.y=id++; sort(begin(f),end(f),greater<pii>()); vector<pair<int,pii> > g(n); id = 0; for(auto &t:g) t.x=rd(); for(auto &t:g) t.y.x=rd(),t.y.y=++id; sort(begin(g),end(g),greater<pair<int,pii>>()); vector<int> ans(m); auto chk = [&](int x) { auto now = g.begin(); auto pos = f.begin(); int sum = 0; priority_queue<pii,vector<pii>,greater<pii>> q; while (pos!=f.end()) { while (now!=g.end()&&now->x>=pos->x) q.push((now++)->y); if (q.empty()) return 0; auto mi = q.top(); q.pop(); if (!mi.y||sum+mi.x>s) return 0; sum += mi.x; int t = x; while (pos!=f.end()&&t--) ans[(pos++)->y]=mi.y; } return 1; }; int l=1,r=m,ret=-1; while (l<=r) { int mid = (l+r)/2; if (chk(mid)) ret=mid,r=mid-1; else l=mid+1; } if (ret==-1) return cout<<"NO"<<endl,0; cout<<"YES"<<endl; chk(ret); for (auto &t:ans) cout<<t<<‘ ‘; cout<<endl; }
377C Captains Mode
大意: $n$个英雄, 给出两个队伍的操作序列, 操作$(p,x)$表示第$x$队选一个英雄, 操作$(b,x)$表示第$x$队禁一个英雄. 一个英雄被选过或被禁过就不能再选, 禁英雄操作可以跳过. 第一个队想要最大化两个队英雄属性和的差, 第二个队想要最小化. 求最优情况下的差值.
只需考虑属性前$m$大的英雄即可, 那么很容易有$O(m^22^m)$的$dp$
但是可以注意到禁英雄一定会比不禁英雄更优, 那么每个阶段可用英雄数是固定的, 所以第$i$个阶段状态数是$\binom{m}{m-i}$, 这样复杂度可以达到$O(m2^m)$.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head const int N = 2e6+10; int n, m, a[N], f[N]; int dp[N]; void chkmin(int &a, int b) {a>b?a=b:0;} void chkmax(int &a, int b) {a<b?a=b:0;} int main() { scanf("%d", &n); REP(i,0,n-1) scanf("%d",a+i); scanf("%d", &m); sort(a,a+n,greater<int>()); REP(i,0,m-1) { char op; int x; scanf(" %c%d", &op, &x); f[i] = (op==‘p‘)<<1|(x==1); } int mx = (1<<m)-1; REP(S,1,mx) { int i = m-__builtin_popcount(S); int &r = dp[S]; if (f[i]==0) { r = INF; REP(j,0,m-1) if (S>>j&1) { chkmin(r,dp[S^1<<j]); } } else if (f[i]==1) { r = -INF; REP(j,0,m-1) if (S>>j&1) { chkmax(r,dp[S^1<<j]); } } else if (f[i]==2) { r = INF; REP(j,0,n-1) if (S>>j&1) { chkmin(r,dp[S^1<<j]-a[j]); } } else if (f[i]==3) { r = -INF; REP(j,0,n-1) if (S>>j&1) { chkmax(r,dp[S^1<<j]+a[j]); } } } printf("%d\n",dp[mx]); }
377D Developing Game
大意: $n$个工人, 第$i$个工人属性$v_i$, 只能和属性范围$[L_i,R_i]$的人一起工作, 求最多选出多少工人.
这道题还是挺有意思的, 刚开始想了一个$dp$的做法, 用树套树优化, 但是数据范围太大了, 很难卡过.
实际上可以直接考虑最终确定的区间$[L,R]$, 那么一个工人$(L_i,v_i,R_i)$会对所有$L_i\le L\le v_i,v_i\le R\le R_i$的区间产生贡献. 这样就转化为二维区间加, 最终查询所有点最值, 可以用线段树扫描线很容易实现.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head const int N = 1e6+10; int n, cnt, mx; struct { int l,v,r; } e[N]; struct _ { int l,r,h,v; } a[N]; struct node { int mx,tag,pos; void upd(int x) {mx+=x,tag+=x;} node operator + (const node &rhs) const { node ret; ret.mx = max(mx,rhs.mx); ret.pos = ret.mx==mx?pos:rhs.pos; ret.tag = 0; return ret; } } tr[N<<2]; void pd(int o) { if (tr[o].tag) { tr[lc].upd(tr[o].tag); tr[rc].upd(tr[o].tag); tr[o].tag=0; } } void add(int o, int l, int r, int ql, int qr, int v) { if (ql<=l&&r<=qr) return tr[o].upd(v); pd(o); if (mid>=ql) add(ls,ql,qr,v); if (mid<qr) add(rs,ql,qr,v); tr[o] = tr[lc]+tr[rc]; } void build(int o, int l, int r) { tr[o].mx=tr[o].tag=0,tr[o].pos=l; if (l!=r) build(ls),build(rs); } int main() { cin>>n; REP(i,1,n) { cin>>e[i].l>>e[i].v>>e[i].r; a[++cnt] = {e[i].l,e[i].v,e[i].v,1}; a[++cnt] = {e[i].l,e[i].v,e[i].r+1,-1}; mx = max(mx, e[i].r+1); } sort(a+1,a+1+cnt,[](_ a,_ b){return a.h<b.h;}); build(1,1,mx); int now = 1, ans = 0; pii pos; REP(i,1,mx) { while (now<=cnt&&a[now].h<=i) add(1,1,mx,a[now].l,a[now].r,a[now].v),++now; if (tr[1].mx>ans) { ans = tr[1].mx; pos = pii(tr[1].pos,i); } } printf("%d\n", ans); REP(i,1,n) { if (e[i].l<=pos.x&&pos.x<=e[i].v&&e[i].v<=pos.y&&pos.y<=e[i].r) { printf("%d ",i); } } puts(""); }
377E Cookie Clicker
大意: $n$种建筑, 第$i$个花费$c_i$个曲奇, 每秒生产$v_i$曲奇. 每秒钟只能选一个建筑工作. 初始时刻$0$, 曲奇数$0$, 若$t$时刻选择一个建筑工作, 那么$t+1$时刻得到收益. 求得到$s$块曲奇最少用时.
斜率优化dp
include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head const int N = 1e6+10; int n; ll s, dp[N]; struct _ { ll c, v; bool operator < (const _ &rhs) const { return c<rhs.c||c==rhs.c&&v<rhs.v; } } a[N]; ll slope2(int i, int j) { return (a[j].v-a[i].v-1+dp[i]-dp[j])/(a[j].v-a[i].v); } ll chk(int i, int j, int k) { return slope2(i,j)>=slope2(j,k); } ll slope(int i, int j) { return (dp[i]-dp[j])/(a[j].v-a[i].v); } int main() { cin>>n>>s; REP(i,1,n) cin>>a[i].v>>a[i].c; sort(a+1,a+1+n); memset(dp,INF,sizeof dp); deque<int> q; ll ans = 1e18; REP(i,1,n) { if (q.size()&&a[q.back()].v>=a[i].v) continue; while (q.size()>1&&slope(q[0],q[1])*a[q[0]].v+dp[q[0]]<a[i].c) q.pop_front(); if (i==1) dp[i] = 0; else if (q.size()) { ll t = (a[i].c-dp[q[0]]+a[q[0]].v-1)/a[q[0]].v; ll rem = t*a[q[0]].v+dp[q[0]]-a[i].c; dp[i] = rem-t*a[i].v; } else continue; ans = min(ans, (s-dp[i]+a[i].v-1)/a[i].v); while (q.size()>1&&chk(q[q.size()-2],q.back(),i)) q.pop_back(); q.push_back(i); } printf("%lld\n", ans); }
Codeforces Round #222 (Div. 1) (ABCDE)
标签：不能 and += inf view ring air tin begin
原文地址：https://www.cnblogs.com/uid001/p/11298911.html
