P4001 [ICPC-Beijing 2006]狼抓兔子

题目地址

易错点：

• 必须熟练掌握当无法在该点继续流量时直接剪枝（d[x]=0）的操作.
• 无向图的最大流由于两边都可增广，应当全部设置为相同的容量.
• 特殊矩阵图的构造.

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int MAXN=1000010,MAXM=1000010;
struct Edge{
	int from,to,w,nxt;
}e[MAXM*6];
int head[MAXN*6],edgeCnt=1;
void addEdge(int u,int v,int w){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].w=w;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
int s,t;
int d[MAXN*6];
bool bfs(){
	memset(d,0,sizeof(d));
	queue<int> q;
	d[s]=1;
	q.push(s);
	while(!q.empty()){
		int nowV=q.front();q.pop();
		for(int i=head[nowV];i;i=e[i].nxt){
			int nowNode=e[i].to;
			if(e[i].w&&(!d[nowNode])){
				d[nowNode]=d[nowV]+1;
				if(nowNode==t)return 1;
				q.push(nowNode);
			}
		}
	}
	return 0;
}
int Dinic(int x,int flow){
	if(x==t)return flow;
	int rest=flow;
	for(int i=head[x];i&&rest;i=e[i].nxt){
		int nowV=e[i].to;
		if(d[nowV]==d[x]+1&&e[i].w){
			int k=Dinic(nowV,min(rest,e[i].w));
			if(!k)d[nowV]=0;
			e[i].w-=k,e[i^1].w+=k;
			rest-=k;
		}
	}
	return flow-rest;
}
int m;
int getHash(int i,int j){
	return (i-1)*m+j;
}
const int INF=2e9;
int main(){
	int n;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m-1;j++){
			int u,v,w;
			scanf("%d",&w);
			u=getHash(i,j),v=getHash(i,j+1);
			addEdge(u,v,w);
			addEdge(v,u,w);
		}
	}
	for(int i=1;i<=n-1;i++){
		for(int j=1;j<=m;j++){
			int u,v,w;
			scanf("%d",&w);
			u=getHash(i,j),v=getHash(i+1,j);
			addEdge(u,v,w);
			addEdge(v,u,w);
		}
	}
	for(int i=1;i<=n-1;i++){
		for(int j=1;j<=m-1;j++){
			int u,v,w;
			scanf("%d",&w);
			u=getHash(i,j),v=getHash(i+1,j+1);
			addEdge(u,v,w);
			addEdge(v,u,w);
		}
	}
	s=1,t=getHash(n,m);
	int ans=0;
	while(bfs()){
		ans+=Dinic(s,INF);
	}
	printf("%d\n",ans);
	return 0;
}

标签：Beijing,head,return,edgeCnt,int,2006,P4001,nowV,nowNode
来源： https://blog.csdn.net/u012972031/article/details/95946679