●BZOJ 3931 [CQOI2015]网络吞吐量

题链：

http://www.lydsy.com/JudgeOnline/problem.php?id=3931

题解：

在最短路图上跑网络流，要开long long
（无奈 BZOJ AC 不了，洛谷上 wa 了一个点。改不出来了诶）

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
struct Edge{
	ll to[500000],cap[500000],nxt[500000],head[2000],ent;
	void Init(){ent=2;}
	void Adde(ll u,ll v,ll w){
		to[ent]=v; cap[ent]=w; nxt[ent]=head[u]; head[u]=ent++;
	}
	ll Next(ll i,bool type){
		return type?head[i]:nxt[i];
	}
}E1,E2;
ll dis[2000],d[2000],cur[2000];
ll N,M,S,T;
ll idx(ll u,ll k){
	return u+N*k;
}
struct cmp{
	bool operator ()(ll a,ll b){
		return dis[a]>dis[b];
	}
};
void Dijkstra(){
	static bool vis[2000];
	priority_queue<ll,vector<ll>,cmp>q;
	memset(dis,0x3f,sizeof(dis));
	dis[1]=0; q.push(1); ll u,v,w;
	while(!q.empty()){
		u=q.top(); q.pop();
		if(vis[u]) continue; vis[u]=1;
		for(ll i=E1.Next(u,1);i;i=E1.Next(i,0)){
			v=E1.to[i],w=E1.cap[i];
			if(dis[v]<=dis[u]+w) continue;
			dis[v]=dis[u]+w;
			q.push(v);
		}
	}
}
bool bfs(){
	queue<ll> q;
	memset(d,0,sizeof(d));
	d[S]=1; q.push(S); ll u,v;
	while(!q.empty()){
		u=q.front(); q.pop();
		for(ll i=E2.Next(u,1);i;i=E2.Next(i,0)){
			v=E2.to[i];
			if(d[v]||!E2.cap[i]) continue;
			d[v]=d[u]+1; q.push(v);
		}
	}
	return d[T];
}
ll dfs(ll u,ll reflow){
	if(u==T||!reflow) return reflow;
	ll flowout=0,f,v;
	for(ll &i=cur[u];i;i=E2.Next(i,0)){
		v=E2.to[i];
		if(d[v]!=d[u]+1) continue;
		f=dfs(v,min(reflow,E2.cap[i]));
		flowout+=f; E2.cap[i^1]+=f;
		reflow-=f;	E2.cap[i]-=f;
		if(!reflow) break;
	}
	if(!flowout) d[u]=0;
	return flowout;
}
ll Dinic(){
	ll flow=0;
	while(bfs()){
		memcpy(cur,E2.head,sizeof(E2.head));
		flow+=dfs(S,INF);
	}
	return flow;
}
int main()
{
	E1.Init(); E2.Init();
	scanf("%lld%lld",&N,&M);S=idx(1,1); T=idx(N,0);
	for(ll i=1,u,v,w;i<=M;i++){
		scanf("%lld%lld%lld",&u,&v,&w);
		E1.Adde(u,v,w);	E1.Adde(v,u,w);
	}
	Dijkstra();
	for(ll u=1;u<=N;u++)
		for(ll i=E1.Next(u,1);i;i=E1.Next(i,0)){
			ll v=E1.to[i],w=E1.cap[i];
			if(dis[u]+w!=dis[v]) continue;
			E2.Adde(idx(u,1),idx(v,0),INF);
			E2.Adde(idx(v,0),idx(u,1),0);
		}
	for(ll i=1,x;i<=N;i++){
		scanf("%lld",&x);
		E2.Adde(idx(i,0),idx(i,1),x);
		E2.Adde(idx(i,1),idx(i,0),0);
	}
	ll ans=Dinic();
	printf("%lld",ans);
	return 0;
}