hdu4496 D-City 判断两元素是否在同一集合

Luxer is a really bad guy. He destroys everything he met.
One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants to know how many connected blocks of D-city left after Luxer destroying the first K D-lines in the input.
Two points are in the same connected blocks if and only if they connect to each other directly or indirectly.

Input

First line of the input contains two integers N and M.
Then following M lines each containing 2 space-separated integers u and v, which denotes an D-line.
Constraints:
0 < N <= 10000
0 < M <= 100000
0 <= u, v < N.

Output

Output M lines, the ith line is the answer after deleting the first i edges in the input.

Sample Input

5 10 0 1 1 2 1 3 1 4 0 2 2 3 0 4 0 3 3 4 2 4

Sample Output

1

1

1

2

2

2

2

3

4

5

思路：从输入的m条边倒着数，
若删完，结果应为n。
从下往上，添加边，若两点不在一个集合中，结果--；

注意：题目没说是一组还是多组用例，这就很尴尬了。
经测验是多组用例。。。。
还有注意数组越界问题，题目n，m范围，以及城市编号是从0到n-1

#include <cstdio>#include <algorithm>#define N 10005#define M 100050using namespace std;int n, m;int a[M], b[M];int ans[M];int f[N];void init(){for (int i = 0; i < n; i++)f[i] = i;}int find(int x){return x == f[x] ? x : f[x] = find(f[x]);}int merge(int x, int y){int t1, t2;t1 = find(x); t2 = find(y);if (t1 != t2){f[t2] = t1;return 1;}return 0;}int main(){while (scanf("%d%d", &n, &m) != EOF){for (int i = 0; i < m; i++)scanf("%d%d", &a[i], &b[i]);init();int tmp = n;for (int i = m - 1; i >= 0; i--){ans[i] = tmp;if (find(a[i]) != find(b[i]))//当前两城镇不连通tmp--;merge(a[i], b[i]);//连接两城镇}for (int i = 0; i < m; i++)printf("%d\n", ans[i]);}return 0;}