ACM: 二分图最佳匹配 图论题 poj 3…

                                        The Windy's

Description

The Windy's is a world famous toy factory that owns Mtop-class workshop to make toys. This year the manager receivesN orders for toys. The manager knows that every order willtake different amount of hours in different workshops. Moreprecisely, the i-th order will take Z_ijhours 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 hasfinished the previous one. The switch does notcost any time.

The manager wants to minimize the average of the finishing timeof the N orders. Can you help him?

Input

The first line of input is the number of test case. The firstline of each test case contains two integers, N and M(1 ≤ N,M ≤ 50).
The next N lines each contain M integers, describingthe 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. Theresult 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

2.0000001.0000001.333333题意: 玩具公司现在有M台机器, N个任务, 要你求出完成每个任务的平均时间最少.

解题思路:

       1. 一看题目的时候一头雾水, 看别人的题解: 二分图最佳匹配问题. 想起KM算法.

       2. 现在将二分图分成X,Y两部分. X部有N个点, Y部有N*M个点

       3. 将每台机器拆成N个点,使得Y部有N*M个点, 

            (1)第J个机器的第P个点代表: 使用J机器进行倒数第P次加工.

            (2)假设N个订单执行的时间分别是t1,t2,t3,...,tn. 执行n个订单的执行时间是:

                t1*n + t2*(n-1) + t3*(n-2) + ... + tn-1*2 + tn

            (3)将每个机器J拆成N个点,第k个点表示倒数第k个订单在此机器上完成.

               连边的权值为: graph[i][j]*k 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 105
#define _clr(x) memset(x,0xff,sizeof(int)*n)
const int INF = (1<<29);

int graph[MAX][MAX];
int mat[MAX][MAX*MAX];
int match1[MAX*MAX], match2[MAX*MAX];
int n, m;

int kuhn_munkras(int m,int n,int mat[][MAX*MAX],int *match1,int *match2)
{
 int s[MAX*MAX], t[MAX*MAX], l1[MAX*MAX], l2[MAX*MAX];
 int p, q, i, j, k;
 int ret = 0;
 for(i = 0; i < m; ++i)
 {
  l1[i] = -INF;
  for(j = 0; j < n; ++j)
  {
   l1[i] = mat[i][j] > l1[i] ? mat[i][j] : l1[i];
  }
 }
 for(i = 0; i < n; ++i) l2[i] = 0;

 _clr(match1), _clr(match2);
 for(i = 0; i < m; ++i)
 {
  for(_clr(t),s[p=q=0] = i; p <= q && match1[i] < 0; ++p)
  {
   for(k = s[p], j = 0; j < n && match1[i] < 0; ++j)
   {
    if(l1[k]+l2[j] == mat[k][j] && t[j] < 0)
    {
     s[++q] = match2[j];
     t[j] = k;
     if(s[q] < 0)
     {
      for(p = j; p >= 0; j = p)
      {
       match2[j] = k = t[j];
       p = match1[k];
       match1[k] = j;
      }
     }
    }
   }
  }
  if(match1[i] < 0)
  {
   for(i--, p = INF, k = 0; k <= q; ++k)
    for(j = 0; j < n; ++j)
     if(t[j] < 0 && l1[s[k]] + l2[j] - mat[s[k]][j] < p)
      p = l1[s[k]] + l2[j] - mat[s[k]][j];
   for(j = 0; j < n; l2[j] += t[j] < 0 ? 0 : p, j++);
   for(k = 0; k <= q; l1[s[k++]] -= p);
  }
 }
 for(i = 0; i < m; ++i)
  ret += mat[i][match1[i]];
 return ret;
}

int main()
{
// freopen("input.txt","r",stdin);
 int i, j;
 int caseNum;
 scanf("%d",&caseNum);
 while(caseNum--)
 {
  scanf("%d %d",&n,&m);
  for(i = 0; i < n; ++i)
  {
   for(j = 0; j < m; ++j)
    scanf("%d",&graph[i][j]);
  }

  for(i = 0; i < n; ++i)
  {
   for(j = 0; j < m; ++j)
   {
    for(int k = 0; k < n; ++k)
    {
     mat[i][j*n+k] = -graph[i][j]*(k+1);
    }
   }
  }
  int result = -kuhn_munkras(n,n*m,mat,match1,match2);
  printf("%.6lf\n",(double)result/n);
 }
 
 return 0;
}