[JSOI 2008] 星球大战

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=1015

[算法]

          考虑离线 ， 将删点转化为加点 ， 用并查集维护连通性即可

          时间复杂度 : O(NlogN)
[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 400010
#define MAXM 200010

struct edge
{
        int to,nxt;
} e[MAXM << 1];

int n,m,cnt,tot,k;
int a[MAXN],ans[MAXN],head[MAXN],atk[MAXN],fa[MAXN];
bool alive[MAXN];

template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; 
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}
inline int get_root(int x)
{
        if (fa[x] == x) return x;
        return fa[x] = get_root(fa[x]);
}
inline void merge(int x,int y)
{
        int u = get_root(x) , v = get_root(y);
        fa[u] = v;
}

int main() 
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++) fa[i] = i;
        for (int i = 1; i <= m; i++)
        {
                int x , y;
                scanf("%d%d",&x,&y);
                x++; y++;
                addedge(x,y);    
                addedge(y,x);
        }
        memset(alive,true,sizeof(alive));
        scanf("%d",&k);
        for (int i = 1; i <= k; i++)
        {
                scanf("%d",&atk[i]);
                alive[++atk[i]] = false;        
        }
        for (int u = 1; u <= n; u++)
        {
                for (int i = head[u]; i; i = e[i].nxt)
                {
                        int v = e[i].to;
                        if (alive[u] && alive[v])
                                merge(u,v);
                }
        }
        for (int i = 1; i <= n; i++) 
        {
                if (alive[i] && get_root(i) == i)
                        cnt++;        
        }
        ans[k + 1] = cnt;
        for (int i = k; i >= 1; i--)
        {
                alive[atk[i]] = true;
                cnt++;
                for (int x = head[atk[i]]; x; x = e[x].nxt)
                {
                        int v = e[x].to;
                        if (alive[v] && get_root(atk[i]) != get_root(v))
                        {
                                merge(atk[i],v);
                                cnt--;
                        }
                }        
                ans[i] = cnt;
        }
        for (int i = 1; i <= k + 1; i++) printf("%d
",ans[i]);
        
        return 0;
    
}