zoukankan      html  css  js  c++  java
  • Hdoj 1102.Constructing Roads 题解

    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


    思路

    就是求最小生成树的问题,已经修过的路就把距离设置为0就好了,用的是Kruskal算法

    代码

    #include<bits/stdc++.h>
    using namespace std;
    int father[110];
    struct Graph
    {
    	int u;
    	int v;
    	int dis;
    }maps[10010];
    
    int grid[110][110];
    void init(int n)
    {
        for(int i=1;i<=n;i++) father[i]=i;
    }
    int find(int x)
    {
        while(father[x]!=x) x=father[x];
        return x;
    }
    void join(int a,int b)
    {
        int t1=find(a);
        int t2=find(b);
        if(t1!=t2) father[t1]=t2;
    }
    int getNum(int n)
    {
    	int num = 0;
    	for(int i=1;i<=n;i++)
    		if(father[i]==i)
    			num++;
    	return num;
    }
    int main()
    {
    	int n;
    	while(cin>>n)
    	{
    		memset(grid,0,sizeof(grid));
    		for(int i=1;i<=n;i++)
    			for(int j=1;j<=n;j++)
    				cin >> grid[i][j];
    		int q;
    		cin >> q;
    		while(q--)
    		{
    			int a,b;
    			cin >> a >> b;
    			grid[a][b] = 0;
    			grid[b][a] = 0;
    		}
    		
    		int edgeNum = 0;
    		for(int i=1;i<=n;i++)
    			for(int j=1;j<=n;j++)
    			{
    				if(i==j)continue;
    				maps[++edgeNum].u = i;
    				maps[edgeNum].v = j;
    				maps[edgeNum].dis = grid[i][j];
    			}
    		sort(maps+1,maps+1+edgeNum,[](Graph x,Graph y)->bool{ return x.dis<y.dis;});
    		init(n);
    		int distance = 0;
    		for(int i=1;i<=edgeNum;i++)
    		{
    			if(find(maps[i].u) != find(maps[i].v))
    			{
    				join(maps[i].u,maps[i].v);
    				distance += maps[i].dis;
    			}
    		}
    		cout << distance << endl;
    	}
    	return 0;
    }
    
  • 相关阅读:
    SQL利用Case When Then多条件判断
    SQL 中LTrim、RTrim与Trim的用法
    SELECT 与 SET 对变量赋值的区别(存储过程)
    exec/sp_executesql语法
    SQLServer : EXEC和sp_executesql的区别
    使用系统监视器监视系统性能
    Centos6.5下OpenLdap搭建(环境配置+双机主从配置+LDAPS+enable SSHA)
    迁移与裁剪linux系统
    Windows Server 2008 R2远程桌面服务配置和授权激活(转)
    cookie、session、sessionid 与jsessionid(转)
  • 原文地址:https://www.cnblogs.com/MartinLwx/p/10042340.html
Copyright © 2011-2022 走看看