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): 13995    Accepted Submission(s): 6690


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

 

分析

给出一张图,求从1到n的最大流。

本来很简单的最大流问题,结果刚好写了一道最小费用最大流的题目,所以直接拿来代码用啦!

 

AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;

const int MAXX = 1005;
const int INF = 0x3f3f3f3f;

struct edge
{
    int to;     //边终点
    int next;   //下一个兄弟位置
    int cap;    //容量
    int flow;   //流量
    int cost;   //费用
} edge[MAXX<<1];

int head[MAXX],tol;
int pre[MAXX],dis[MAXX];
int N;  //节点总个数

void init(int n)
{
    N=n;
    tol=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int cap,int cost)  //同时增加原边与反向边
{
    edge[tol].to=v;
    edge[tol].cap=cap;
    edge[tol].cost=cost;
    edge[tol].flow=0;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].cost=-cost;
    edge[tol].flow=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}

/*
 * SPFA 算法判断是否存在s到t的通路
 */
bool spfa(int s,int t)
{
    queue<int>q;
    bool vis[MAXX];
    for(int i=0; i<N; i++)
    {
        dis[i]=INF;
        vis[i]=false;
        pre[i]=-1;
    }
    dis[s]=0;
    vis[s]=true;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=false;
        for(int i=head[u]; i!=-1; i=edge[i].next)   //遍历所有与u临接的点
        {
            int v=edge[i].to;
            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)    //如果可以松弛该点
            {
                dis[v]=dis[u]+edge[i].cost;
                pre[v]=i;
                if(!vis[v]) //如果该点不在队列中,入队
                {
                    vis[v]=true;
                    q.push(v);
                }
            }
        }
    }
    return (pre[t]!=-1);    //返回是否s到t是否有路径
}

/*
 * int s 起点
 * int t 终点
 * 返回最大流flow
 */
int minCostMaxFlow(int s,int t)
{
    int flow=0;
    while(spfa(s,t))
    {
        int minn=INF;   //当前路径可增广值
        for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])   //因为建图时每增加一条边会同时加入它的反向边,因此i^1为找出与i刚好相反的部分
        {
            if(minn>edge[i].cap-edge[i].flow)
                minn=edge[i].cap-edge[i].flow;
        }
        for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])   //修改图,计算花费
        {
            edge[i].flow+=minn;
            edge[i^1].flow-=minn;
        }
        flow+=minn;
    }
    return flow;
}

int main()
{
    int T;
    scanf("%d",&T);
    for(int ti=1; ti<=T; ti++)
    {
        int n,m,s,t,cost;
        scanf("%d%d",&n,&m);
        init(n+1);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&s,&t,&cost);
            addedge(s,t,cost,1);
        }
        printf("Case %d: %d\n",ti,minCostMaxFlow(1,n));
    }
    return 0;
}