大意: $n$天, 第$i$天要$a_i$个志愿者. $m$种志愿者, 每种无限多, 第$i$种工作时间$[s_i,t_i]$花费$c_i$, 求最少花费.
源点$S$连第一天, 容量$INF$
第$n+1$天连汇点$T$, 容量$INF$
第$i$天往后连$INF-a_i$
每个志愿者连$s_i$到$t_i+1$, 容量$INF$, 费用$c_i$
求出$S$到$T$的最小费用最大流即可
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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; const int N = 1e6+10, INF = 0x3f3f3f3f, S = N-2, T = N-1; int n, m, flow; struct edge { int to,w,v,next; edge(int to=0,int w=0,int v=0,int next=0):to(to),w(w),v(v),next(next){} } e[N]; int head[N], dep[N], vis[N], cur[N], f[N], cnt=1; int pre[N],pre2[N]; queue<int> Q; ll cost; int spfa() { REP(i,1,n+1) f[i]=dep[i]=INF,vis[i]=0; f[S]=dep[S]=f[T]=dep[T]=INF; dep[S]=0,Q.push(S); while (Q.size()) { int u = Q.front(); Q.pop(); vis[u] = 0; for (int i=head[u]; i; i=e[i].next) { if (dep[e[i].to]>dep[u]+e[i].v&&e[i].w) { dep[e[i].to]=dep[u]+e[i].v; pre[e[i].to]=u,pre2[e[i].to]=i; f[e[i].to]=min(f[u],e[i].w); if (!vis[e[i].to]) { vis[e[i].to]=1; Q.push(e[i].to); } } } } return dep[T]!=INF; } void EK(){ while(spfa()) { int w = f[T]; for (int u=T; u!=S; u=pre[u]) { e[pre2[u]].w-=w; e[pre2[u]^1].w+=w; } flow += w, cost += (ll)w*dep[T]; } } void add(int u, int v, int w, int k) { e[++cnt] = edge(v,w,k,head[u]); head[u] = cnt; e[++cnt] = edge(u,0,-k,head[v]); head[v] = cnt; } int main() { scanf("%d%d",&n,&m); REP(i,1,n) { int t; scanf("%d",&t); add(i,i+1,INF-t,0); } add(S,1,INF,0); add(n+1,T,INF,0); REP(i,1,m) { int s,t,c; scanf("%d%d%d",&s,&t,&c); add(s,t+1,INF,c); } EK(); printf("%lld\n",cost); }