POJ 3686 The Windy's 最小费用最大流

每个工厂拆成N个工厂，费用分别为1~N倍原费用。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt","r",stdin);
   // freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
 
const int maxn = ;
int N, M;
int cost[maxn][maxn];
 
void Input()
{
    scanf("%d%d", &N, &M);
    for(int i = ; i <= N; i++)
        for(int j = ; j <= M; j++)
            scanf("%d", &cost[i][j]);
}
 
struct Edge
{
    int from, to, cost, cap;
};
const int maxv = maxn * maxn + maxn;
vector<int> g[maxv];
vector<Edge> edge;
int n,s,t;
 
void add(int from, int to, int cost, int cap)
{
    edge.push_back((Edge){from, to, cost, cap});
    g[from].push_back(edge.size() - );
    edge.push_back((Edge){to, from, -cost, });
    g[to].push_back(edge.size() - );
}
 
void init()
{
    for(int i = ; i <= n; i++)
        g[i].clear();
    edge.clear();
}
 
void construct()
{
    n = N + N * M + ;
    s = n - ;
    t = n;
    init();
 
    for(int i = ; i <= N; i++)
        add(s, i, , );
    for(int k = ; k <= N; k++)
    {
        for(int j = ; j <= M; j++)
        {
            int id = N + (k - ) * M + j;
            add(id, t, , );
            for(int i = ; i <= N; i++)
                add(i, id, k * cost[i][j], );
        }
    }
}
 
int d[maxv];
int inq[maxv];
int road[maxv];
int SPFA()
{
    memset(d, INF, sizeof(d));
    memset(inq, , sizeof(inq));
    queue<int> q;
    q.push(s);
    d[s] = ;
    road[s] = -;
 
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        inq[u] = false;
 
        for(int i = ; i < g[u].size(); i++)
        {
            Edge &e = edge[g[u][i]];
            if(e.cap >  && d[u] + e.cost < d[e.to])
            {
                d[e.to] = d[u] + e.cost;
                road[e.to] = g[u][i];
                if(!inq[e.to])
                {
                    inq[e.to] = true;
                    q.push(e.to);
                }
            }
        }
    }
    return d[t] != INF;
}
int MCMF()
{
    int cost = ;
    while(SPFA())
    {
        cost += d[t];
        for(int e = road[t]; e != -; e = road[edge[e].from])
        {
            edge[e].cap -= ;
            edge[e^].cap += ;
        }
    }
    return cost;
}
int main()
{
    debug();
    int t;
    scanf("%d", &t);
    for(int ca = ; ca <= t; ca++)
    {
        Input();
        construct();
        printf("%.6f\n", MCMF() * 1.0 / N);
    }
    return ;
}