BNU - 4216 - 修路 （并查集判断连通分量）
修路
Time Limit: 1000ms
Memory Limit: 65536KB
64-bit integer IO format:
%lld Java class name:
Main
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在某个景区内有n个景点，它们之间有m条路相连。然而，这m条路可能是不足够的，因为无法把这n个景点都连通起来。
例如当m<n-1的时候，就必定有部分景点被孤立。
所以，现在政府想修建道路，想把这些景点都连通起来。问题是，在现在的基础上，最少要再修建多少条道路呢？
Input
第一行是n,m (1<=n<=1e5,1<=m<=1e6)
接下来就是m行，每行两个1 ~ n范围内的数，表示编号为这两个数的景点之间有一道路相连
Output
输出最少要再修建的道路的条数
Sample Input
5 2 1 2 3 4
Sample Output
2
Source
第八届北京师范大学程序设计竞赛热身赛第四场
思路：就是去寻找连通分量的个数cnt，最少修建的道路个数即为cnt-1（犯了两超傻逼错误╮(╯_╰)╭）
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 100005;
int pa[maxn], vis[maxn];
int find(int x) {
return pa[x] != x ? pa[x] = find(pa[x]) : x;
}
int main() {
int n, m;
while(scanf("%d %d", &n, &m) != EOF) {
for(int i = 0; i <= n; i++) pa[i] = i;
memset(vis, 0, sizeof(vis));
while(m--) {
int a, b;
scanf("%d %d", &a, &b);
a = find(a), b = find(b);
if(a != b) pa[a] = b;
}
int cnt = 0;
for(int i = 1; i <= n; i++) {
int tmp = find(i);
if(!vis[tmp]) {
vis[tmp] = 1;
cnt++;
}
}
printf("%d\n", cnt-1);
}
return 0;
}