HDU 4496 D-City (并查集)

题意：给定一个图，问你每次删除一条边后有几个连通块。

析：水题，就是并查集的运用，倒着推。

代码如下：

#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int p[maxn];
int e1[maxn*10], e2[maxn*10];
int Find(int x){  return x == p[x] ? x : p[x] = Find(p[x]); }
int ans[maxn*10];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i)  p[i] = i;

        for(int i = 0; i < m; ++i)
            scanf("%d %d", &e1[i], &e2[i]);

        int cnt = n;
        for(int i = m-1; i >= 0; --i){
            ans[i] = cnt;
            int x = Find(e1[i]);
            int y = Find(e2[i]);
            if(x != y){
                p[y] = x;
                --cnt;
            }
        }

        for(int i = 0; i < m; ++i)
            printf("%d
", ans[i]);
    }
    return 0;
}