hdu 3549 Flow Problem【Dinic最大流】

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本文详细介绍了一种经典的计算机科学问题——最大流问题，并提供了一个具体的实现案例。针对多组测试数据，通过输入节点数量及边的数量来构建加权有向图，并利用Dinic算法求解从源点到汇点的最大流值。

Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 12954 Accepted Submission(s): 6175

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

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

Source

HyperHexagon's Summer Gift (Original tasks)

题目大意：

t组数据，n个点，m条边，求最大流。

记录属于自己的模板！

Dinic 358ms

Ford_Fulkerson 1808ms

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
int head[100000];
struct edge
{
    int from,to,w,next;
}e[1000000];
int div[100000];
int n,m,cont,ss,tt;
void add(int from,int to,int w)
{
    e[cont].to=to;
    e[cont].w=w;
    e[cont].next=head[from];
    head[from]=cont++;
}
int makediv()
{
    memset(div,0,sizeof(div));
    queue<int >s;
    s.push(ss);
    div[ss]=1;
    while(!s.empty())
    {
        int u=s.front();
        if(u==tt)return 1;
        s.pop();
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            int w=e[i].w;
            if(w&&div[v]==0)
            {
                div[v]=div[u]+1;
                s.push(v);
            }
        }
    }
    return 0;
}
int Dfs(int u,int maxflow,int tt)
{
    if(u==tt)return maxflow;
    int ret=0;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].to;
        int w=e[i].w;
        if(w&&div[v]==div[u]+1)
        {
            int f=Dfs(v,min(maxflow-ret,w),tt);
            e[i].w-=f;
            e[i^1].w+=f;
            ret+=f;
            if(ret==maxflow)return ret;
        }
    }
    return ret;
}
void Dinic()
{
    ss=1;tt=n;
    int ans=0;
    while(makediv()==1)
    {
        ans+=Dfs(ss,INF,tt);
    }
    printf("%d\n",ans);
}
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        cont=0;
        memset(head,-1,sizeof(head));
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,w);
            add(y,x,0);
        }
        printf("Case %d: ",++kase);
        Dinic();
    }
}