[poj3686]The Windy's(费用流)

题目大意：

解题关键：指派问题，待更。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<vector>
#include<queue>
#define inf 0x3f3f3f3f
#define MAX_V 10010
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
struct edge{int to,cap,cost,rev;};
int V;
vector<edge>G[MAX_V];
int h[MAX_V],dist[MAX_V],prevv[MAX_V],preve[MAX_V];
void add_edge(int from,int to,int cap,int cost){
    G[from].push_back((edge){to,cap,cost,G[to].size()});
    G[to].push_back((edge){from,,-cost,G[from].size()-});
}

int min_cost_flow(int s,int t,int f){
    int res=;
    fill(h,h+V,);
    while(f>){
        priority_queue<P,vector<P>,greater<P> >que;
        fill(dist,dist+V+,inf);
        dist[s]=;
        que.push(P(,s));
        while(!que.empty()){
            P p=que.top();que.pop();
            int v=p.second;
            if(dist[v]<p.first) continue;
            for(int i=;i<G[v].size();i++){
                edge &e=G[v][i];
                if(e.cap>&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                    dist[e.to]=dist[v]+e.cost+h[v]-h[e.to];
                    prevv[e.to]=v;
                    preve[e.to]=i;
                    que.push(P(dist[e.to],e.to));
                }
            }
        }
        if(dist[t]==inf) return -;
        for(int v=;v<V;v++) h[v]+=dist[v];

        //增广
        int d=f;
        for(int v=t;v!=s;v=prevv[v]) d=min(d,G[prevv[v]][preve[v]].cap);
        f-=d;
        res+=d*h[t];
        for(int v=t;v!=s;v=prevv[v]){
            edge &e=G[prevv[v]][preve[v]];
            e.cap-=d;
            G[v][e.rev].cap+=d;
        }
    }
    return res;
}
int n,m,t1,t2,t3,t4;
int z[][];
int main(){
    int _;
    scanf("%d",&_);
    while(_--){
        memset(G,,sizeof G);
        scanf("%d%d",&n,&m);
        for(int i=;i<n;i++){
            for(int j=;j<m;j++){
                scanf("%d",&z[i][j]);
            }
        }
        int s=n+n*m,t=s+;
        V=t+;
        for(int i=;i<n;i++) add_edge(s,i,,);
        for(int j=;j<m;j++){
            for(int k=;k<n;k++){
                add_edge(n+j*n+k,t,,);
                for(int i=;i<n;i++){
                    add_edge(i,n+j*n+k,,(k+)*z[i][j]);
                }
            }
        }
        printf("%.6f\n",(double)min_cost_flow(s,t,n)/n);
    }
    return ;
}