ACM刷题之HDU————Constructing Roads

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22339    Accepted Submission(s): 8591

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

 

最小生成树的题目

因为每个点都有连接，所以要找出最小的，通过观察，发现左下角和右上角是对称的

所以我取了左下角的数据进行排序，再用克鲁斯卡尔算法生成最小生成树。

下面是ac代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;

struct nd{
	int x,y,val;
};

int a[105][105];
int cnt;
nd w[5051];
int pre[105];
int b[105][105]; 

bool cmp(const nd &a,const nd &b){
	return a.val<b.val;
}

int finds(int u){
	return u==pre[u]?u:pre[u]=finds(pre[u]);
}

void join(int a, int b,int v)
{
	int p1 = finds(a);
	int p2 = finds(b);
	if(p1==p2)	return;
	pre[p2]=p1;
	cnt+=v;
}


int main()
{
	//freopen("f:/input.txt", "r", stdin);
	int i,j,k,n,m,zu,q;
	while(scanf("%d",&zu)!=EOF)
	{
		CLR(a,0);
		CLR(w,0);
		CLR(b,0);
		k=0;
		cnt = 0;
		for(i=0;i<zu;i++)
		{
			for(j=0;j<zu;j++)
				scanf("%d",&a[i][j]);
		}
		
		scanf("%d",&q);
		while(q--)
		{
			scanf("%d%d",&n,&m);
			b[n-1][m-1]=b[m-1][n-1]=-1;
		}
		
		
		for(i=1,j=0;i<zu;i++)
		{
			for(j=0;j<i;j++)
			{
				w[k].x = i;
				w[k].y = j;
				if(b[i][j]==-1)
					w[k].val = 0;
				else
					w[k].val = a[i][j];
				++k;
			}
				
		}
		for(i=0;i<=100;i++)
			pre[i]=i;
		
		
		sort(w,w+k,cmp);
		for(i=0;i<k;i++)
		{
			join(w[i].y,w[i].x,w[i].val);
		}
		
		printf("%d\n",cnt);
		
	} 
	
	
}