hdu1102：Constructing Roads（最小生成树）

最新推荐文章于 2020-06-02 10:49:09 发布

原创最新推荐文章于 2020-06-02 10:49:09 发布 · 167 阅读

0 ·

CC 4.0 BY-SA版权

hdu 同时被 3 个专栏收录

96 篇文章

订阅专栏

hdu刷题记录

72 篇文章

订阅专栏

最小生成树

18 篇文章

订阅专栏

本文深入探讨了如何利用最小生成树算法解决村庄间道路连接问题，通过实例详细讲解了算法的实现过程，包括输入村庄数量、距离矩阵及已建道路信息，采用Prim算法求解最优路径，最终输出连接所有村庄所需的最小总成本。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://acm.hdu.edu.cn/showproblem.php?pid=1102

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

Sample Output

题意分析：

给出所有村庄之间的道路修建费用，有点路已经修建，求连接所有村庄需要最少的花费。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 120
int e[N][N], book[N], dis[N];
int main()
{
	int n, i, j, t, sum, u, v, mini, count, inf=99999999;
	while(scanf("%d", &n)!=EOF)
	{
		for(i=1; i<=n; i++)
			for(j=1; j<=n; j++)
				scanf("%d", &e[i][j]);
		memset(book, 0, sizeof(book));
		scanf("%d", &t);	
		while(t--)
		{
			scanf("%d%d", &u, &v);
			e[u][v]=e[v][u]=0;
		}
		sum=0;
		for(i=1; i<=n; i++)
			dis[i]=e[1][i];
		
		book[1]=1;
		count=1;
		while(count<n)
		{
			mini=inf;
			for(i=1; i<=n; i++)
			{
				if(dis[i]<mini && book[i]==0)
				{
					mini=dis[i];
					u=i;
				}
			}
			sum+=dis[u];
			book[u]=1;
			count++;
			for(v=1; v<=n; v++)
			{
				if(dis[v]>e[u][v] && book[v]==0)	
					dis[v]=e[u][v];
			}
		}
		printf("%d\n", sum);
	}
	return 0;
}