POJ 3686 —— 最小费用流||最小费用匹配&KM

The Windy's Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 3520 Accepted: 1509 Description The Windy's is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receives N orders for toys. The manager knows that every order will take different amount of hours in different workshops. More precisely, the i-th order will take Zij hours if the toys are making in the j-th workshop. Moreover, each order's work must be wholly completed in the same workshop. And a workshop can not switch to another order until it has finished the previous one. The switch does not cost any time. The manager wants to minimize the average of the finishing time of the N orders. Can you help him? Input The first line of input is the number of test case. The first line of each test case contains two integers, N and M (1 ≤ N,M ≤ 50). The next N lines each contain M integers, describing the matrix Zij (1 ≤ Zij ≤ 100,000) There is a blank line before each test case. Output For each test case output the answer on a single line. The result should be rounded to six decimal places. Sample Input 3 3 4 100 100 100 1 99 99 99 1 98 98 98 1 3 4 1 100 100 100 99 1 99 99 98 98 1 98 3 4 1 100 100 100 1 99 99 99 98 1 98 98 Sample Output 2.000000 1.000000 1.333333 Source POJ Founder Monthly Contest – 2008.08.31, windy7926778

题意是有n个玩具，交给m个工厂，j号工厂加工i号玩具需要Z[i][j]时间，求加工完所有玩具的平均时间。

思路：这题有两种解法比较方便，首先是最小费用匹配KM算法，据说极快；第二种是最小费用流，时间860MS，比较慢。我用的是第二种。

我们把多个工厂加工一个玩具转化为一个多个只能加工一种玩具的多个工厂，他们分别花费1- N倍的时间。然后就是一个经典的指派问题，最小费用流可解。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 50 + 5;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int MOD = (int)1e9 + 7;
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int , int> pi;
const int mod = 1000000000;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int t , n , m ;
int Z[MAXN][MAXN];
struct edge{int to , cap , cost ,rev;};
vector <edge> G[MAXN * MAXN];
int V;
int h[MAXN * MAXN];
int dist[MAXN * MAXN];
int prevv[MAXN  * MAXN] , preve[MAXN*MAXN];
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 , 0 , -cost , G[from].size() - 1});
}
int min_cost_flow(int s , int t , int f)
{
    int res = 0;
    fill(h , h + V , 0);
    while(f > 0)
    {
        priority_queue<pi , vector<pi> , greater<pi> >que;
        fill(dist , dist + V , INF);
        dist[s] = 0;
        que.push(pi(0 , s));
        while(!que.empty())
        {
            pi p = que.top();que.pop();
            int v = p.second;
            if(dist[v] < p.first)continue;
            for(int i = 0 ; i < G[v].size() ; i++)
            {
                edge &e = G[v][i];
                if(e.cap > 0 && 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(pi(dist[e.to] , e.to));
                }
            }
        }
        if(dist[t] == INF)return -1;
        for(int v = 0 ; v <V ; v++)h[v] += dist[v];
        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;
}
void solve()
{
    ///0 -- n - 1：玩具
    ///n -- 2* n - 1:0号工厂
    ///2 * n -- 3 * n - 1： 1号工厂
    ///...
    ///m * n -- （m + 1） * n - 1：m-1号工厂
    int s = n * m + n , t= s + 1;
    V = t + 1;
    for(int i = 0 ; i < n ; i++)
    {
        add_edge(s , i , 1 , 0);
    }
    for(int j = 0 ; j < m ; j++)
    {
        for(int k = 0 ; k < n ; k++)
        {
            add_edge(n + j * n + k , t , 1 , 0);
            for(int i = 0; i < n ; i++)
            {
                add_edge(i , n + j * n + k , 1 , (k + 1) * Z[i][j]);
            }
        }
    }
    printf("%.6f\n" , (double)min_cost_flow(s , t , n) / n);
}

int main()
{
    cin >> t;
    while(t--)
    {
        clr(preve , 0);
        clr(prevv , 0);
        clr(G , 0);
        scanf("%d%d" , &n , &m);
        for(int i = 0 ; i < n ; i++)
        {
            for(int j = 0 ; j < m ; j++)
            {
                scanf("%d",&Z[i][j]);
            }
        }
        solve();
    }
    return 0;
}