Hdu 3549 Flow Problem(最大流)

Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 14412 Accepted Submission(s): 6859
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
Sample Output
Case 1: 1
Case 2: 2
Author
HyperHexagon

/* 裸题. 求1~n的最大流. 多组数据哦. */
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#define MAXN 10001
using namespace std;
struct data{int v,next,c;}e[MAXN*2];
int n,m,tot,cut,s=1,max1=1e9,head[MAXN],dis[MAXN];
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
    return x*f;
}
void add(int u,int v,int x)
{
    e[++cut].v=v;
    e[cut].c=x;
    e[cut].next=head[u];
    head[u]=cut;
}
int bfs()
{
    queue<int >q;
    memset(dis,-1,sizeof dis);
    dis[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];i;i=e[i].next)
        {
            int v=e[i].v;
            if(dis[v]==-1&&e[i].c>0)
            {
                dis[v]=dis[u]+1;
                q.push(v);
            }
        }
    }
    return dis[n]!=-1;
}
int dfs(int u,int y)
{
    int rest=0;
    if(u==n) return y;
    for(int i=head[u];i&&rest<y;i=e[i].next)
    {
        int v=e[i].v;
        if(e[i].c>0&&dis[v]==dis[u]+1)
        {
            int x=min(e[i].c,y-rest);
            x=dfs(v,x);
            rest+=x;
            e[i].c-=x;
            e[i^1].c+=x;
        }
    }
    if(!rest) dis[u]=-1;
    return rest;
}
int dinic(int s,int t)
{
    int tot=0;
    while(bfs())
    {
        int ans=dfs(s,max1);
        while(ans){tot+=ans;ans=dfs(s,max1);}
    }
    return tot;
}
void Clear()
{
    cut=1;//2WA
    memset(dis,-1,sizeof dis);
    memset(head,0,sizeof head);
}
int main()
{
    int t,u,v,x;t=read();
    for(int i=1;i<=t;i++)
    {
        n=read();m=read();
        Clear();
        while(m--)
        {
            u=read(),v=read(),x=read();
            add(u,v,x),add(v,u,0);//1WA
        }
        printf("Case %d: %d\n",i,dinic(1,n));
    }
    return 0;
}