洛谷 P1197 [JSOI2008]星球大战

题目传送门

解题思路：

并查集，逆序做（即先假设给的k个星球全都被炸，求出此时的联通块个数，就是经过k次打击的联通块个数。然后再加上最后一个被炸的星球，就求出了经过k-1次打击的联通块个数。。。以此类推，最后把所有点都加进去，就求出了经过0次打击后连同块个数）

//转载自：https://www.luogu.org/blog/user25199/solution-p1197

AC代码：

#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>

using namespace std;

int n,m,fa[400001],tot,br[400001],sum[400002],head[400001],k;
struct kkk {
    int from,next,to;
}g[400001];
bool vis[400001];

void add(int a,int b) {
    g[++tot].from = a;
    g[tot].to = b;
    g[tot].next = head[a];
    head[a] = tot; 
}

int find_father(int x) {
    if(fa[x] == x) return x;
    return fa[x] = find_father(fa[x]);
}

void merge(int x,int y) {
    int x1 = find_father(x);
    int y1 = find_father(y);
    fa[y1] = x1;
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 0;i <= n; i++) {
        fa[i] = i;
        head[i] = -1;
    }
    for(int i = 1;i <= m; i++) {
        int x,y;
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }
    scanf("%d",&k);
    for(int i = 1;i <= k; i++) {
        scanf("%d",&br[i]);
        vis[br[i]] = 1;
    }
    int ans = n - k;
    for(int i = 1;i <= 2 * m; i++) {
        if(!vis[g[i].from] && !vis[g[i].to] && find_father(g[i].from) != find_father(g[i].to)){
            ans--;
            merge(g[i].from,g[i].to);
        }        
    }
    sum[k+1] = ans;
    for(int i = k;i >= 1; i--) {
        ans++;
        vis[br[i]] = 0;
        for(int j = head[br[i]];j != -1;j = g[j].next) {
            if(!vis[g[j].to] && find_father(br[i]) != find_father(g[j].to)) {
                ans--;
                merge(br[i],g[j].to);
            }
        }
        sum[i] = ans;
    }
    for(int i = 1;i <= k + 1; i++)
        printf("%d
",sum[i]);
    return 0;
}