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维护联通块自然想到并查集，然而题中说是删边，不是很好做，因此我们可以离线下来然后倒序操作，就变成了添加边的同时维护联通块数量。

首先我们把k次打击后剩的边都添加到图中，表示倒序时的初始状态。然后将 i 从 k 到1枚举，将第 i 个被袭击的星球 del[i] 连的所有边都加入图中，同时维护并查集，当然要满足他连的星球也是未被袭击的用一个bool数组维护即可。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<queue>
  8 #include<stack>
  9 #include<vector>
 10 #include<cctype>
 11 using namespace std;
 12 #define enter puts("")
 13 #define space putchar(' ')
 14 #define Mem(a) memset(a, 0, sizeof(a))
 15 typedef long long ll;
 16 typedef double db;
 17 const int INF = 0x3f3f3f3f;
 18 const db eps = 1e-8;
 19 const int maxn = 2e6  + 5;
 20 inline ll read()
 21 {
 22     ll ans = 0;
 23     char ch = getchar(), last = ' ';
 24     while(!isdigit(ch)) {last = ch; ch = getchar();}
 25     while(isdigit(ch))
 26     {
 27         ans = ans * 10 + ch - '0'; ch = getchar();
 28     }
 29     if(last == '-') ans = -ans;
 30     return ans;
 31 }
 32 inline void write(ll x)
 33 {
 34     if(x < 0) putchar('-'), x = -x;
 35     if(x >= 10) write(x / 10);
 36     putchar(x % 10 + '0');
 37 }
 38 
 39 int n, m, k;
 40 vector<int> v[maxn << 1];
 41 struct Node
 42 {
 43     int x, y;
 44 }t[maxn];
 45 int del[maxn];
 46 bool vis[maxn << 1];
 47 int num;
 48 
 49 int p[maxn << 1];
 50 void init(int n)
 51 {
 52     for(int i = 1; i <= n; ++i) p[i] = i;
 53 }
 54 int Find(int x)
 55 {
 56     return x == p[x] ? x : p[x] = Find(p[x]);
 57 }
 58 
 59 int ans[maxn];
 60 
 61 int main()
 62 {
 63     n = read(); m = read();
 64     init(n);
 65     for(int i = 1; i <= m; ++i) 
 66     {
 67         int x = read(), y = read();
 68         v[x].push_back(y); v[y].push_back(x);
 69     }
 70     k = read();
 71     for(int i = 1; i <= k; ++i)
 72     {
 73         del[i] = read();
 74         vis[del[i]] = 1;
 75     }
 76     num = n - k;
 77     for(int i = 1; i <= n; ++i) if(!vis[i])
 78     {
 79         for(int j = 0; j < (int)v[i].size(); ++j)
 80         {
 81             if(!vis[v[i][j]])
 82             {
 83                 int px = Find(i), py = Find(v[i][j]);
 84                 if(px != py) {p[px] = py; num--;}    //要合并成一个联通块 
 85             }
 86         }
 87     }
 88     for(int i = k; i > 0; --i)
 89     {
 90         ans[i] = num;
 91         for(int j = 0; j < (int)v[del[i]].size(); ++j)
 92         {
 93             int e = v[del[i]][j];
 94             if(!vis[e])
 95             {
 96                 int px = Find(del[i]), py = Find(e);
 97                 if(px != py) {p[px] = py; num--;}
 98             }
 99         }
100         vis[del[i]] = 0; num++;        //别忘了，现在del[i]变成了未被袭击的星球 
101     }
102     ans[0] = num;
103     for(int i = 0; i <= k; ++i) {write(ans[i]); enter;}
104     return 0;
105 }

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