HDU 3435 A new Graph Game

题意：给出一张有n个点m条边带权的无向图，要你找出若干个回路并且使得所有的点都在某个哈密顿回路上，且所有的点只能出现在一个哈密顿回路中，求由n条边组成的若干回路的最小权值。

跟HDU 3488基本一样，只不过把有向图改成了无向图。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define maxn 2010
#define maxm 410000
#define INF 1<<30
using namespace std;
struct MCMF{
    int src,sink,e,n;
    int first[maxn];
    int cap[maxm],cost[maxm],v[maxm],next[maxm];
    bool flag;
    void init(){
        e = 0;
        memset(first,-1,sizeof(first));
    }

    void add_edge(int a,int b,int cc,int ww){
        cap[e] = cc;cost[e] = ww;v[e] = b;
        next[e] = first[a];first[a] = e++;
        cap[e] = 0;cost[e] = -ww;v[e] = a;
        next[e] = first[b];first[b] = e++;
    }

    int d[maxn],pre[maxn],pos[maxn];
    bool vis[maxn];

    bool spfa(int s,int t){
        memset(pre,-1,sizeof(pre));
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        for(int i = 0;i <= n;i++)   d[i] = INF;
        Q.push(s);pre[s] = s;d[s] = 0;vis[s] = 1;
        while(!Q.empty()){
            int u = Q.front();Q.pop();
            vis[u] = 0;
            for(int i = first[u];i != -1;i = next[i]){
                if(cap[i] > 0 && d[u] + cost[i] < d[v[i]]){
                    d[v[i]] = d[u] + cost[i];
                    pre[v[i]] = u;pos[v[i]] = i;
                    if(!vis[v[i]])  vis[v[i]] = 1,Q.push(v[i]);
                }
            }
        }
        return pre[t] != -1;
    }

    int Mincost;
    int Maxflow;

    int MinCostFlow(int s,int t,int nn){
        Mincost = 0,Maxflow = 0,n = nn;
        while(spfa(s,t)){
            int min_f = INF;
            for(int i = t;i != s;i = pre[i])
                if(cap[pos[i]] < min_f) min_f = cap[pos[i]];
            Mincost += d[t] * min_f;
            Maxflow += min_f;
            for(int i = t;i != s;i = pre[i]){
                cap[pos[i]] -= min_f;
                cap[pos[i]^1] += min_f;
            }
        }
        return Mincost;
    }
};
MCMF g;

int main()
{
    int n,m,t;
    scanf("%d",&t);
    for(int kase = 1;kase <= t;kase++){
        scanf("%d%d",&n,&m);
        g.init();
        int S = 0,T = 2*n+1;
        for(int i = 1;i <= n;i++)
            g.add_edge(S,i,1,0),g.add_edge(i+n,T,1,0);
        for(int i = 0;i < m;i++){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g.add_edge(a,b+n,1,c);
            g.add_edge(b,a+n,1,c);
        }
        int ans = g.MinCostFlow(S,T,T);
        if(g.Maxflow == n)
            printf("Case %d: %d
",kase,ans);
        else
            printf("Case %d: NO
",kase);
    }
    return 0;
}