【POJ 3686 The Windy's】+ 最小费用流

The Windy’s
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 5412 Accepted: 2268
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

题意 ： n 个玩具 ，m 个工厂，j 号工厂加工 i 号玩具需要 a[i][j] 的时间，没个玩具必须在一个工厂内完成，玩具的制作顺序可以是任意的，每个工厂每次只能加工一个玩具，求加工完所有的玩具的平均最小时间

思路 : 一个工厂时 T = a[1] + (a[1] + a[2]) + ……+ (a[1] + a[2] + …..a[n]）
变形得 ： T = n * a[1] + （n - 1） * a[2] + ……+ a[n];
同样也可以看成多个只能制作一个玩具得工厂 + 最小费用流

AC代码：

#include<cstdio>#include<vector>#include<algorithm>using namespace std;const int MAX = 3e3 + 10;const int INF = 0x3f3f3f3f;int a[52][52],n,m,V;int dis[MAX]; // 最短距离int pre[MAX],pr[MAX];// 最短路中的前驱结点和对应的边struct edge{ int to,cap,cost,rev; };vector <edge> G[MAX];void add(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;    while(f > 0){        fill(dis,dis + V,INF);        dis[s] = 0;        bool update = true;        while(update){            update = false;            for(int v = 0; v < V; v++){                if(dis[v] == INF) continue;                for(int i = 0; i < G[v].size(); i++){                    edge &e = G[v][i];                    if(e.cap > 0 && dis[e.to] > dis[v] + e.cost){                        dis[e.to] = dis[v] + e.cost;                        pre[e.to] = v;                        pr[e.to] = i;                        update = true;                    }                }            }        }        if(dis[t] == INF) return -1;        int d = f;        for(int v = t; v != s; v = pre[v])            d = min(d,G[pre[v]][pr[v]].cap);        f -= d;        res += d * dis[t];        for(int v = t; v != s; v = pre[v]){            edge &e = G[pre[v]][pr[v]];            e.cap -= d;            G[v][e.rev].cap += d;        }    }    return res;}void solve(){    int s = n + n * m,t = s + 1;    V = t + 1;    for(int i = 0; i < MAX; i++)        G[i].clear();    for(int i = 0; i < n; i++)        add(s,i,1,0);    for(int j = 0; j < m; j++)        for(int k = 0; k < n; k++){            add(n + j * n + k,t,1,0);            for(int i = 0; i < n; i++)                add(i,n + j * n + k,1,(k + 1) * a[i][j]);        }    printf("%.6f\n",(double) min_cost_flow(s,t,n) / n);}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d %d",&n,&m);        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                scanf("%d",&a[i][j]);        solve();    }    return 0;}