poj 3686

The Windy's

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3791		Accepted: 1631

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 Z_ij 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 Z_ij (1 ≤ Z_ij ≤ 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

 

每个玩具点对（1 ~ N) * z[i][j]费用的 工厂 建边，求最小费用流

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int MAX = 60;
const int INF = 1e9 + 7;
int N, M;
int z[MAX][MAX];
struct Edge {int from, to, cap, flow, cost;};
vector<Edge> edges;
vector<int> G[MAX * MAX * MAX + 3 * MAX];
int p[MAX * MAX * MAX + 3 * MAX], a[MAX * MAX * MAX + 3 * MAX];
int d[MAX * MAX * MAX + 3 * MAX];
bool inq[MAX * MAX * MAX + 3 * MAX];

void add_edge(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});
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
}

bool bellmanford(int s, int t, int &flow, int &cost) {
        for(int i = s; i <= t; ++i) {
                d[i] = INF;
        }

        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

        queue<int> Q;
        Q.push(s);
        while(!Q.empty()) {
                //printf("fuck\n");
                int u = Q.front(); Q.pop();
                inq[u] = 0;
                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;
                                p[e.to] = G[u][i];
                                a[e.to] = min(a[u], e.cap - e.flow);
                                if(!inq[e.to]) {
                                        inq[e.to] = 1;
                                        Q.push(e.to);
                                }
                        }
                }
        }

        if(d[t] == INF) return 0;
        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 1;
}

int Mincost(int s, int t) {
        int flow = 0, cost = 0;
        while(bellmanford(s, t, flow, cost));
        return cost;
}

int main()
{
    //freopen("sw.in", "r", stdin);
    int t;
    scanf("%d", &t);
    while(t--) {
            scanf("%d%d", &N, &M);
            for(int i = 1; i <= N; ++i) {
                    for(int j = 1; j <= M; ++j) {
                            scanf("%d", &z[i][j]);
                            //printf("%d ", z[i][j]);
                    }
            }

            int s = 0, t = N + N * M + 1;
            for(int i = s; i <= t; ++i) G[i].clear();
            edges.clear();

            for(int i = 1; i <= N; ++i) {
                    add_edge(s, i, 1, 0);
            }


                    for(int j = 1; j <= M; ++j) {
                            for(int k = 1; k <= N; ++k) {
                                    add_edge(j * N + k, t, 1, 0);
                                    for(int i = 1; i <= N; ++i) {
                                            add_edge(i, j * N + k, 1, z[i][j] * k);
                                    }
                            }
                    }
                    printf("%.6f\n", (double) Mincost(s, t) / N);


    }

    //cout << "Hello world!" << endl;
    return 0;
}