poj-2421 Constructing Roads（最小生成树 Kruskal算法）

Constructing Roads

Time Limit:2000MS     Memory Limit:65536KB    

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

题意：n个村庄，及村庄之间的距离，q条路已连通村庄，问连通所有村庄的最短路径。

Kruskal算法的思想：

按照边权值递增的顺序排序边集e；

建立由n棵树组成的森林，每棵树包含图的一个节点；

最小生成树的权和sum=0；

for(k=1;k<=m;k++)  //枚举e中的所有边

{ if (第k条边(i ,j)的两个短点分属于两颗子树)

{ i节点所在子树并入j节点所在子树；

sum+=(i ,j)的边长 }

}

输出sum；

把每棵树看成一个集合，变成了典型的并查集运算。

#include
#include
#include
using namespace std;
struct node
{
	int i,j;
	int distant;
}f[10000];
int s[101];
int find(int x)
{
	return s[x]==x?x:s[x]=find(s[x]);
}
bool cmp(const node &a,const node &b)
{
	return a.distant