HDU 3549 Flow Problem(有向边网络流)

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cctype>
#include <queue>
#include <stdlib.h>
#include <cstdlib>
#include <math.h>
#include <set>
#include <vector>
#define inf 100000000
#define eps 1e-8
#define N 205
#define M 1050
#define ll int
using namespace std;
inline ll Max(ll a,ll b){return a>b?a:b;}
inline ll Min(ll a,ll b){return a<b?a:b;}
//M为边数 N为点数 从1-n
//M为边数 N为点数 从1-n
struct Edge{
	int from,to,flow,cap, nex;
}edge[M*2];//双向边，注意RE 注意这个模版是 相同起末点的边 合并而不是去重
int head[N],edgenum;//2个要初始化-1和0
void addedge(int u,int v,int cap){//网络流要加反向弧
	Edge E={u,v,0,cap,head[u]};
	edge[edgenum]=E;
	head[u]=edgenum++;
	Edge E2={v,u,0,0,head[v]}; //这里的cap若是单向边要为0
	edge[edgenum]=E2;
	head[v]=edgenum++;
}


int dis[N],cur[N];//距离起点的距离 cur[i]表示i点正在考虑的边 优化不再考虑已经用过的点 初始化为head
bool vis[N];
bool BFS(int Start,int End){
	memset(vis,0,sizeof(vis)); 
	memset(dis,-1,sizeof(dis));
	queue<int>Q;	while(!Q.empty())Q.pop();
	Q.push(Start);	dis[Start]=0;	vis[Start]=1;
	while(!Q.empty())
	{
		int u = Q.front(); Q.pop();
		for(int i=head[u];i!=-1;i=edge[i].nex){
			Edge E =edge[i];
			if(!vis[E.to] && E.cap>E.flow)
			{
				vis[E.to]=1;
				dis[E.to]=dis[u]+1;
				if(E.to==End)return true;
				Q.push(E.to);
			}
		}
	}
	return false;
}
int DFS(int x, int a,int End){//流入x 的流量是a
	if(x==End || a==0)return a;
	int flow = 0, f;
	for(int& i=cur[x];i!=-1;i=edge[i].nex)
	{
		Edge& E = edge[i];
		if(dis[x]+1 == dis[E.to] && (f = DFS(E.to , Min(a, E.cap-E.flow), End))>0 )
		{
			E.flow += f;
			edge[ i^1 ].flow -= f;//反向边要减掉
			flow += f;
			a -= f;
			if(a==0)break;
		}
	}
	return flow;
}
int Maxflow(int Start,int End){
	int flow=0; 
	while(BFS(Start,End)){
		memcpy(cur,head,sizeof(head));//把head的数组复制过去
		flow += DFS(Start, inf, End);
	}
	return flow;
}
int main() {
	int T,Cas=1,n,m,i,a,b,c;scanf("%d",&T);
	while (T--) {
		memset(head,-1,sizeof(head));
		edgenum=0;
		scanf("%d %d",&n,&m);		
		while(m--)
		{
			scanf("%d %d %d",&a,&b,&c);
			addedge(a,b,c);
		}
		printf("Case %d: %d\n",Cas++,Maxflow(1,n));
	}
	return 0;
}

/*
99
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1

3 2
1 3 2
1 3 5

3 2
1 2 456
1 2 56431

3 3 
1 3 100
1 1 100
1 1 100

11 15
1 2 1 
1 3 1
1 4 1
2 5 1
2 6 1
3 7 1
3 8 1
4 9 1
4 10 1
5 11 1
6 11 1
7 11 1
8 11 1
9 11 1
10 11 1

11 15
1 2 2
1 3 2
1 4 2
2 5 1
2 6 1
3 7 1
3 8 1
4 9 1
4 10 1
5 11 1
6 11 1
7 11 1
8 11 1
9 11 1
10 11 1

11 15
1 2 2
1 3 1
1 4 2
2 5 1
2 6 1
3 7 1
3 8 1
4 9 1
4 10 1
5 11 1
6 11 1
7 11 1
8 11 1
9 11 1
10 11 1

2 0

2 1
1 2 4651

4 4
1 2 10
2 1 5
2 4 20
1 3 3

4 5
1 2 10
2 1 5
2 4 20
1 3 3
3 4 1

9 10
1 5 2
2 4 6
2 3 4 
1 2 9
3 9 5
5 9 4
2 3 1
4 2 1
6 7 1
3 7 2

4 4
1 2 10
1 3 2
3 4 1
2 4 1

4 4
1 2 1
1 3 1
3 4 10
2 4 10

ans:
1
2
7
0
100
3
6
5
0
4651
10
11
7
2
2

*/