HDU 3435 A new Graph Game(费用流)

HDU 3435 A new Graph Game(费用流)

http://acm.hdu.edu.cn/showproblem.php?pid=3435

题意:

给你一个N个节点M条边的无向图,要你求该图有1个或多个不相交有向环(哈密顿回路)构成时,所有这些有向环的最小权值.

分析:

本题之前用KM算法做过:

http://blog.csdn.net/u013480600/article/details/38780451

要注意,可以从本题的第3个用例的输出可以看出,本题的无向边,其实就是等效于两条方向相反的有向边.(如果本题的无向边==一条有向边,那么用例3无解). 所以对于本题来说无向图其实就是有向图的所有边必须添加两边(仅此而已),本题与之前所做HDU1853几乎一样的解法,详细可以参考HDU1853的分析:

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

建图如下(论证参考HDU1853):

源点s编号0,n个节点编号1到n和n+1到2*n, 汇点t编号2*n+1.

从源点s到第i个点有边 (s, i, 1, 0)

如果有无向边(i,j,w),那么有两条有向边 (i, j+n, 1, w) 和 (j, i+n, 1, w)

每个节点i到汇点t有边 (i+n, t, 1, 0)

最终我们求得最大流如果==n的话,最小费用就是解. 否则输出NO.

AC代码: C++3890MS, G++3671MS

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define INF 1e9
using namespace std;
const int maxn = 2000+5;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int from,int to,int cap,int flow,int cost):from(from),to(to),cap(cap),flow(flow),cost(cost){}
};

struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    int d[maxn];
    int a[maxn];
    int p[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;++i) G[i].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));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int &flow,int &cost)
    {
        for(int i=0;i<n;++i) d[i]=INF;
        queue<int> Q;
        memset(inq,0,sizeof(inq));
        d[s]=0,inq[s]=true,Q.push(s),a[s]=INF,p[s]=0;

        while(!Q.empty())
        {
            int u=Q.front(); Q.pop();
            inq[u]=false;
            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;
                    a[e.to] = min(a[u],e.cap-e.flow);
                    p[e.to] = G[u][i];
                    if(!inq[e.to]){inq[e.to]=true; Q.push(e.to);}
                }
            }
        }
        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;
    }

    int solve(int num)
    {
        int flow=0,cost=0;
        while(BellmanFord(flow,cost));
        return flow ==num ? cost:-1;
    }
}MM;

int dist[maxn][maxn];

int main()
{
    int T; scanf("%d",&T);
    for(int kase=1;kase<=T;++kase)
    {
        int n,m,src,dst;
        scanf("%d%d",&n,&m);
        src=0, dst=2*n+1;
        MM.init(2*n+2,src,dst);
        memset(dist,-1,sizeof(dist));
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(dist[u][v]==-1 || dist[u][v]>w)
            {
                dist[u][v]=dist[v][u]=w;
            }
        }
        for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j)if(dist[i][j]!=-1)
        {
            MM.AddEdge(i,j+n,1,dist[i][j]);
            MM.AddEdge(j,i+n,1,dist[i][j]);
        }
        for(int i=1;i<=n;++i)
        {
            MM.AddEdge(src,i,1,0);
            MM.AddEdge(i+n,dst,1,0);
        }
        int ans=MM.solve(n);
        if(ans==-1) printf("Case %d: NO\n",kase);
        else printf("Case %d: %d\n",kase,ans);
    }
    return 0;
}