【POJ 2400】 Supervisor, Supervisee（KM求最小权匹配）

【POJ 2400】 Supervisor, Supervisee（KM求最小权匹配）

Description

Input

Output

Sample Input

2
7
1 2 3 4 5 6 7
2 1 3 4 5 6 7
3 1 2 4 5 6 7
4 1 2 3 5 6 7
5 1 2 3 4 6 7
6 1 2 3 4 5 7
7 1 2 3 4 5 6
1 2 3 4 5 6 7
2 1 3 4 5 6 7
3 1 2 4 5 6 7
4 1 2 3 5 6 7
5 1 2 3 4 6 7
6 1 2 3 4 5 7
7 1 2 3 4 5 6

2
1 2
2 1
1 2
1 2

Sample Output

Data Set 1, Best average difference: 0.000000
Best Pairing 1
Supervisor 1 with Employee 1
Supervisor 2 with Employee 2
Supervisor 3 with Employee 3
Supervisor 4 with Employee 4
Supervisor 5 with Employee 5
Supervisor 6 with Employee 6
Supervisor 7 with Employee 7

Data Set 2, Best average difference: 0.250000
Best Pairing 1
Supervisor 1 with Employee 1
Supervisor 2 with Employee 2

Source

题目大意就是n个上司与n名员工。每一个上司相应有想要搭配的员工。相同每一个员工有渴望搭配的上司。

输入第一行为N 之后n行为1~n号上司的期望 从左到右从最好到最差

相同之后n行是1~n号员工

匹配到最渴望的人值为0，否则从左到右一次加1

要求问平均期望的最小值，也就是最小值/2n

最小值用KM最小权匹配计算就可以，因为还要求输出解，有多解则输出多解。所以还要搜一下……

事实上数据非常少。找最小权匹配也用搜的也能够。

。

。

代码例如以下：

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const double eps = 1e-8;

int mp[23][33];
int lx[33],ly[33],link[33],slack[33],next[33];
bool visx[33],visy[33],vis[33];
int n,ans,cnt;

bool cal(int x)
{
	visx[x] = 1;

	for(int y = 0; y < n; ++y)
	{
		if(visy[y]) continue;
		
		int t = lx[x]+ly[y]-mp[x][y];
		if(t == 0)
		{
			visy[y] = 1;
			if(link[y] == -1 || cal(link[y]))
			{
				link[y] = x;
				return 1;
			}
		}
		else slack[y] = min(slack[y],t);
	}
	return 0;
}

int KM()
{
	memset(link,-1,sizeof(link));

	for(int i = 0; i < n; ++i)
	{
		memset(slack,INF,sizeof(slack));
		while(1)
		{
			memset(visx,0,sizeof(visx));
			memset(visy,0,sizeof(visy));

			if(cal(i)) break;

			int d = INF;
			for(int i = 0; i < n; ++i)
				if(!visy[i]) d = min(d,slack[i]);

			for(int i = 0; i < n; ++i)
				if(visx[i]) lx[i] -= d;

			for(int i = 0; i < n; ++i)
				if(visy[i]) ly[i] += d;
				else slack[i] -= d;
		}
	}

	ans = 0;
	for(int i = 0; i < n; ++i)
		if(link[i] != -1) ans += mp[link[i]][i];

	return -ans;
}

void dfs(int id,int hs)
{
	if(hs < ans) return;
	if(id == n)
	{
		if(hs == ans)
		{
			printf("Best Pairing %d\n",++cnt);
			for(int i = 0; i < n; ++i)
			{
				printf("Supervisor %d with Employee %d\n",i+1,next[i]+1);
			}
		}
		return;
	}

	for(int i = 0; i < n; ++i)
	{
		if(vis[i]) continue;
		vis[i] = 1;
		next[id] = i;
		dfs(id+1,hs+mp[id][i]);
		vis[i] = 0;
	}
}

int main()
{
	int t,x;
	scanf("%d",&t);
	
	for(int z = 1; z <= t; ++z)
	{
		scanf("%d",&n);
		memset(ly,0,sizeof(ly));

	
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
			{
				scanf("%d",&x);
				mp[x-1][i] = -j;
			}

		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
			{
				scanf("%d",&x);
				mp[i][x-1] -= j;
				if(j == 0) lx[i] = mp[i][x-1];
				else lx[i] = max(lx[i],mp[i][x-1]);
			}

		printf("Data Set %d, Best average difference: %.6f\n",z,KM()*0.5/n);

		cnt = 0;
		memset(vis,0,sizeof(vis));
		dfs(0,0);
		puts("");
	}

	return 0;
}