HDU 3435A new Graph Game(网络流之最小费用流)

题目地址:HDU 3435

这题刚上来一看,感觉毫无头绪。。再仔细想想。。发现跟我做的前两道费用流的题是差不多的。可以往那上面转换。

建图基本差不多,只不过这里是无向图。建图依然是拆点,判断入度出度,最后判断是否满流,满流的话这时的费用流是符合要求的,输出,不能满流的话,输出NO。

代码如下:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
int head[3000], source, sink, cnt, flow, cost, num;
int d[3000], vis[3000], pre[3000], cur[3000];
queue<int>q;
struct node
{
    int u, v, cap, cost, next;
}edge[10000000];
void add(int u, int v, int cap, int cost)
{
    edge[cnt].v=v;
    edge[cnt].cap=cap;
    edge[cnt].cost=cost;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cap=0;
    edge[cnt].cost=-cost;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int spfa()
{
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));
    int minflow=INF, i;
    q.push(source);
    d[source]=0;
    cur[source]=-1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(d[v]>d[u]+edge[i].cost&&edge[i].cap)
            {
                d[v]=d[u]+edge[i].cost;
                minflow=min(minflow,edge[i].cap);
                cur[v]=i;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
    if(d[sink]==INF) return 0;
    flow+=minflow;
    cost+=minflow*d[sink];
    for(i=cur[sink];i!=-1;i=cur[edge[i^1].v])
    {
        edge[i].cap-=minflow;
        edge[i^1].cap+=minflow;
    }
    return 1;
}
void mcmf(int n)
{
    while(spfa());
        if(flow==n)
        printf("%d\n",cost);
        else
            printf("NO\n");
}
int main()
{
    int t, n, m, i, j, a, b, c;
    scanf("%d",&t);
    num=0;
    while(t--)
    {
        num++;
        scanf("%d%d",&n,&m);
        memset(head,-1,sizeof(head));
        cnt=0;
        source=0;
        sink=2*n+1;
        flow=0;
        cost=0;
        for(i=1;i<=n;i++)
        {
            add(source,i,1,0);
            add(i+n,sink,1,0);
        }
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b+n,1,c);
            add(b,a+n,1,c);
        }
        printf("Case %d: ",num);
        mcmf(n);
    }
    return 0;
}


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