HDU 1102 Constructing Roads (最小生成树,kruskal)

本文介绍了一道经典最小生成树问题——HDU1102 Constructing Roads,并提供了一种有效的求解算法。该算法通过剔除已存在的路径,在剩余的边中构建最小生成树来连接所有村庄。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

HDU 1102 Constructing Roads

题目链接

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24371 Accepted Submission(s): 9395

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
3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output
179

Source
kicc

========================================================================================

依然是一道最小生成树的变形算法,让你在最小生成树里剔除掉一直的部分

下面是AC代码

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;

int mark[110];
int pic[110][110];

struct Node{
    int x,y;
    int weight;
}node[20000];

int find(int x)
{
    return mark[x] == x ? x : mark[x] = find(mark[x]);
}


void join (int x,int y)
{
    int r1 = find(x);
    int r2 = find(y);
    if( r1 < r2)
        mark[r2] = r1;
    else if( r2 < r1)
        mark[r1] = r2;
}

bool cmp(Node x, Node y)
{
    return x.weight < y.weight;
}

int main()
{
    int n,m;
    while(scanf("%d",&n) != EOF)
    {
            for( int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                scanf("%d", &pic[i][j]);
            }
        }
        int k = 0;
        for ( int i = 1; i <= n; i++)
        {
            for ( int j = i+1; j <= n; j++)
            {
                node[k].x = i;
                node[k].y = j;
                node[k].weight = pic[i][j];
                k++;
            }
        }
        for( int i = 1; i <= n; i++)
            mark[i] = i;

        scanf("%d",&m);
         for( int i = 0; i < m; i++)
         {
            int a,b;
            scanf("%d%d", &a, &b);
            join(a,b);
         }

         sort( node, node+k, cmp);
         long long ans = 0;
        for( int i = 0; i < k; i++)
        {
            if( find(node[i].x) != find(node[i].y))
            {
                join( node[i].x,node[i].y );
                ans += node[i].weight;
            }
        }
        printf("%lld\n",ans);
    }


    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值