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HDU 3488 Tour

时间:2015-09-25 18:38:32      阅读:154      评论:0      收藏:0      [点我收藏+]

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有向环最小权值覆盖问题

http://blog.csdn.net/u013480600/article/details/39159407

 

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

//设置节点数量
const int maxn=450+10;

const int INF=0x7FFFFFFF;
struct Edge
{
    int from,to,cap,flow,cost;
};
int n,m,len,s,t;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];


void init()
{
    for(int i=0; i<maxn; i++) G[i].clear();
    edges.clear();
}

void Addedge(int from,int to,int cap,int cost)
{
    edges.push_back((Edge)
    {
        from,to,cap,0,cost
    });
    edges.push_back((Edge)
    {
        to,from,0,0,-cost
    });
    len=edges.size();
    G[from].push_back(len-2);
    G[to].push_back(len-1);
}

bool BellmanFord(int s,int t,int &flow,int &cost)
{

    for(int i=0; i<maxn; i++) d[i]=INF;

    memset(inq,0,sizeof(inq));
    memset(p,-1,sizeof(p));

    d[s]=0;
    inq[s]=1;
    p[s]=0;
    a[s]=INF;

    queue<int>Q;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        inq[u]=0;
        for(int i=0; i<G[u].size(); i++)
        {
            Edge& e=edges[G[u][i]];
            if(e.cap>e.flow&&d[e.to]>d[u]+e.cost)
            {
                d[e.to]=d[u]+e.cost;
                p[e.to]=G[u][i];
                a[e.to]=min(a[u],e.cap-e.flow);
                if(!inq[e.to])
                {
                    Q.push(e.to);
                    inq[e.to]=1;
                }
            }
        }
    }
    if(d[t]==INF) return false;
    flow+=a[t];
    cost+=d[t]*a[t];
    int u=t;
    while(u!=s)
    {
        edges[p[u]].flow+=a[t];
        edges[p[u]^1].flow-=a[t];
        u=edges[p[u]].from;
    }
    return true;
}

void Mincost (int s,int t)
{
    int flow=0,cost=0;
    while(BellmanFord(s,t,flow,cost));
    printf("%d\n",cost);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        s=0; t=2*n+1;
        for(int i=1; i<=n; i++) Addedge(s,i,1,0);
        for(int i=1; i<=n; i++) Addedge(i+n,t,1,0);
        for(int i=1; i<=m; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            Addedge(u,v+n,1,w);
        }
        Mincost(s,t);
    }
    return 0;
}

 

HDU 3488 Tour

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原文地址:http://www.cnblogs.com/zufezzt/p/4838899.html

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