ZOJ 2314 Reactor Cooling [无源汇上下界网络流]

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=205, M=5e5+5, INF=1e9;
inline ll read(){
	char c=getchar();ll x=0,f=1;
	while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

int n, m, u, v, c, b, s, t;
int extra[N];
struct edge{int v, c, f, ne, lower;}e[M];
int cnt=1, h[N];
inline void ins(int u, int v, int c, int b=0) { //printf("ins %d %d  %d\n",u,v,c);
	e[++cnt]=(edge){v, c, 0, h[u], b}; h[u]=cnt;
	e[++cnt]=(edge){u, 0, 0, h[v], b}; h[v]=cnt;
}

int q[N], head, tail, d[N], vis[N], cur[N];
bool bfs() {
	memset(vis, 0, sizeof(vis));
	head=tail=1;
	q[tail++]=s; d[s]=0; vis[s]=1;
	while(head!=tail) {
		int u=q[head++];
		for(int i=h[u];i;i=e[i].ne) 
			if(!vis[e[i].v] && e[i].c > e[i].f) {
				vis[e[i].v]=1; d[e[i].v]=d[u]+1;
				q[tail++]=e[i].v;
				if(e[i].v == t) return true;
			}
	}
	return false;
}
int dfs(int u, int a) {
	if(u==t || a==0) return a;
	int flow=0, f;
	for(int &i=cur[u];i;i=e[i].ne) 	
		if(d[e[i].v] == d[u]+1 && (f=dfs(e[i].v, min(e[i].c-e[i].f, a))) >0) {
			flow += f;
			e[i].f += f;
			e[i^1].f -= f;
			a -= f;
			if(a==0) break;
		}
	if(a) d[u]=-1;
	return flow;
}
int dinic() {
	int flow=0;
	while(bfs()) {
		for(int i=s; i<=t; i++) cur[i]=h[i];
		flow+=dfs(s, INF);
	}
	return flow;
}
int main() {
	freopen("in","r",stdin);
	int T=read();
	while(T--) {
		n=read(); m=read(); s=0; t=n+1;
		cnt=1; memset(h,0,sizeof(h)); memset(extra, 0, sizeof(extra));
		for(int i=1; i<=m; i++) 
			u=read(), v=read(), b=read(), c=read(), ins(u, v, c-b, b), extra[u]-=b, extra[v]+=b;
		int sum=0;
		for(int i=1; i<=n; i++) {
			if(extra[i]>0) ins(s, i, extra[i]), sum+=extra[i];
			else ins(i, t, -extra[i]);
		}
		int flow=dinic(); //printf("flow %d %d\n",flow,sum);
		if(flow!=sum) puts("NO");
		else {
			puts("YES");
			for(int i=1; i<=m; i++) printf("%d\n",e[i<<1].f + e[i<<1].lower);
		}
	}
}

posted @ 2017-03-27 23:31  Candy?  阅读(177)  评论(0编辑  收藏  举报