hdu The Windy's二分图最佳匹配问题

The Windy's

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 1

Problem 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

PKU

这个题目的建图很经典啊，n个产品，m个工厂，由于每个工厂可能会被匹配多次，将工厂拆点，每一个工厂拆成n个（共n*m个），当一个工厂被拆成n个的时候，其中第k个表示第i个产品在第j个工厂上是倒数第k个生产的，那么，之后的k-1个产品的等待时间都要加time[i][j]，也就是说如果第i件产品在第j个工厂上是倒数第k的生产的，则有这件产品生产所导致的时间花费为k*time[i][j]（包括生产时间time[i][j]和其后的k-1个产品等待它的时间(k-1)*time[i][j]），这样做正好避免了这么一个问题--第i件产品在第j个工厂上是第k个被生产的所需要的花费怎么算？由于j工厂生产的前k-1个产品还没有确定，所以没有办法这么做，只能倒着算了，把每个工厂的时间花费拆成其上生产商品k(倒数第k个)所花费的时间，还有k之后的k-1个等待它时间。

不知道为什么在HDOJ 上的WEB-DIY里面交不上去这道题目，在POJ上就可以过。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 55
#define inf 1e9
int n,m,t,nn,mm;
int map[60][3000];
int lx[maxn],ly[3000],my[3000],slack;
bool visx[maxn],visy[3000];
void init(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			int tmp;
			scanf("%d",&tmp);
			for(int k=1;k<=n;k++){
				map[i][j*n+k-1]=-tmp*k;
			}
		}
	}nn=n;mm=n*m;
}
bool dfs(int u){
	visx[u]=true;
	for(int v=0;v<mm;v++)if(!visy[v]){
		int t=lx[u]+ly[v]-map[u][v];
		if(!t){
			visy[v]=true;
			if(my[v]==-1 || dfs(my[v])){
				my[v]=u; return true;
			}
		}else if(slack>t) slack=t;
	}return false;
}
double km(){
	double ans=0;
	for(int i=0;i<nn;i++){
		lx[i]=-inf;
		for(int j=0;j<mm;j++){
			if(map[i][j]>lx[i]) lx[i]=map[i][j];
		}
	}
	for(int j=0;j<mm;j++){ly[j]=0;my[j]=-1;}
	for(int i=0;i<nn;i++){
		while(1){
			memset(visx,0,sizeof(visx));
			memset(visy,0,sizeof(visy));
			slack=inf;
			if(dfs(i)) break;
			for(int j=0;j<nn;j++)if(visx[j])lx[j]-=slack;
			for(int j=0;j<mm;j++)if(visy[j])ly[j]+=slack;
		}
	}
	for(int i=0;i<mm;i++){
		if(my[i]!=-1) ans-=map[my[i]][i];
	}return ans/n;
}
int main()
{
	freopen("in.txt","r",stdin);
	scanf("%d",&t);
	while(t--){
		init();
		printf("%.6lf\n",km());
	}
    return 0;
}