POJ-2421 Constructing Roads (最小生成树)

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25593		Accepted: 11207

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

#include 
#include 
using namespace std;
struct edge{
	int from, to, val;
	bool operator < (const edge& x) const{
		return val < x.val;
	}
}e[11111];
int p[11111], a[111][111];
int findset(int x){
	return p[x] == x ? x : p[x] = findset(p[x]);
}
int main(){
	int n, num;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		for(int j = 1; j <= n; ++j){
			scanf("%d", &a[i][j]);
		}
	}
	num = 0;
	for(int i = 1; i <= n; ++i){
		for(int j = i + 1; j <= n; ++j){
			e[++num].val = a[i][j];
			e[num].from = i;
			e[num].to = j;
		}
	}
	sort(e + 1, e + 1 + num);
	for(int i = 1; i <= n; ++i){
		p[i] = i;
	}
	int q, u, v, ans = 0;
	scanf("%d", &q);
	while(q--){
		scanf("%d %d", &u, &v);
		u = findset(u);
		v = findset(v);
		p[u] = v;
	}
	for(int i = 1; i <= num; ++i){
		u = findset(e[i].from);
		v = findset(e[i].to);
		if(u != v){
			p[u] = v;
			ans += e[i].val;
		}
	}
	printf("%d\n", ans);
}

/*
题意：
100个城市，城市之间已经有一些路了，现在需要再建一些使得所有城市联通，问最小代价。

思路：
最小生成树，先处理一下已经建边的集合，然后照常跑最小生成树即可。并查集维护一下。
*/