POJ3216——Repairing Company

最新推荐文章于 2021-02-17 13:26:15 发布

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最新推荐文章于 2021-02-17 13:26:15 发布

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分类专栏： POJ 之旅文章标签：匹配

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本文介绍了一个修复公司的任务分配问题，通过构建最小路径覆盖问题并利用Floyd算法预处理城市区块间的最短路径，最终采用匈牙利算法求解最小路径覆盖数。

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Repairing Company

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 6325		Accepted: 1699

Description

Lily runs a repairing company that services the Q blocks in the city. One day the company receives M repair tasks, the ith of which occurs in block p_i, has a deadline t_i on any repairman’s arrival, which is also its starting time, and takes a single repairman d_i time to finish. Repairmen work alone on all tasks and must finish one task before moving on to another. With a map of the city in hand, Lily want to know the minimum number of repairmen that have to be assign to this day’s tasks.

Input

The input contains multiple test cases. Each test case begins with a line containing Q and M (0 < Q ≤ 20, 0 < M ≤ 200). Then follow Q lines each with Q integers, which represent a Q × Q matrix Δ = {δ_ij}, where δ_ij means a bidirectional road connects the ith and the jth blocks and requires δ_ij time to go from one end to another. If δ_ij = −1, such a road does not exist. The matrix is symmetric and all its diagonal elements are zeroes. Right below the matrix are M lines describing the repairing tasks. The ith of these lines contains p_i, t_i and d_i. Two zeroes on a separate line come after the last test case.

Output

For each test case output one line containing the minimum number of repairmen that have to be assigned.

Sample Input

Sample Output

Source

POJ Monthly--2007.04.01, crazyb0y

尼玛我想到最小顶点覆盖去了，其实是最小路径覆盖，我们把任务看成点，那么对于每个人，若他修好i之后可以修好j，那么在i，j之间连一条边，说明i和j可以由一个人完成，之后把最小路径覆盖求出来就行，但是要用floyd求出块和块之间的最短路

#include<map>      
#include<set>      
#include<list>      
#include<stack>      
#include<queue>      
#include<vector>      
#include<cmath>      
#include<cstdio>      
#include<cstring>      
#include<iostream>      
#include<algorithm>      
      
using namespace std;  

bool vis[210];
int mark[210];
int mat[25][25];
int q, m;

struct node
{
	int pi;
	int ti;
	int di;
}repair[210];

struct node2
{
	int next;
	int to;
}edge[210 * 210];
int head[210];
int tot;

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool dfs(int u)
{
	for(int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if(!vis[v])
		{
			vis[v] = 1;
			if(mark[v] == -1 || dfs(mark[v]))
			{
				mark[v] = u;
				return true;
			}
		} 
	}
	return false;
}

int hungary()
{
	int ans = 0;
	memset( mark, -1, sizeof(mark) );
	for(int i = 1; i <= m; i++)
	{
		memset( vis, 0, sizeof(vis) );
		if(dfs(i))
			ans++;
	}
	return ans;
}

int main()
{
	while(~scanf("%d%d", &q, &m))
	{
		if(!q && !m)
			break;
		memset( head, -1, sizeof(head) );
		tot = 0;
		for(int i = 1; i <= q; i++)
			for(int j = 1; j <= q; j++)
			{
				scanf("%d", &mat[i][j]);
				if(mat[i][j] == -1)
					mat[i][j] = 0x3f3f3f3f;
			}
		for(int k = 1; k <= q; k++)
		{
			for(int i = 1; i <= q; i++)
			{
				for(int j = 1; j <= q; j++)
				{
					mat[i][j] =min(mat[i][j], mat[i][k] + mat[k][j]);
				}
			}
		}
		for(int i = 1; i <= m; i++)
		{
			scanf("%d%d%d", &repair[i].pi, &repair[i].ti, &repair[i].di);
		}
		for(int i = 1; i <= m; i++)
		{
			for(int j = 1; j <= m; j++)
			{
				if(i == j)
					continue;
				int u = repair[i].pi;
				int v = repair[j].pi;
				if(mat[u][v] + repair[i].di + repair[i].ti <= repair[j].ti)
				{
					addedge(i, j);
				}
			}
		}
		int ans = hungary();
		printf("%d\n", m - ans);
	}
	return 0;
}