[JSOI2008]星球大战

正向进行很显然很BT, 所以逆向离线此题是个不错的选择. 使用并查集统计劫难之后的联通状态, 然后逐个添加结点即可.

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <vector>
 using std::vector;
 #define VERTEXMAX 400003
 int set[VERTEXMAX], Des[VERTEXMAX], output[VERTEXMAX];
 bool Vertex[VERTEXMAX];
 vector<int> To[VERTEXMAX];
 void Initialize()
 {
     memset(Vertex, true, sizeof(Vertex));
     for (int i = ; i < VERTEXMAX; ++i)
         set[i] = i;
 }
 int Father(int x)
 {
     if (set[x] == x)
         return x;
     return set[x] = Father(set[x]);
 }
 void Union(int x, int y, int &sum)
 {
     int a_Father = Father(x);
     int b_Father = Father(y);
     if (a_Father != b_Father)
     {
         set[a_Father] = b_Father;
         --sum;
     }
 }
 int main()
 {
     Initialize();
     int V, E, Dn, Block, p = ;
     scanf("%d %d", &V, &E);
     for (int i = , x, y; i < E; To[x].push_back(y), To[y].push_back(x), ++i)
         scanf("%d %d", &x, &y);
     scanf("%d", &Dn);
     p = Dn;
     for (int i = ; i < Dn; Vertex[Des[i]] = false, ++i)
         scanf("%d", &Des[i]);
     Block = V - Dn;
     for (int i = , j; i < V; ++i)
         if (Vertex[i])
             for (j = ; j < To[i].size(); ++j)
                 if (Vertex[To[i][j]])
                     Union(i, To[i][j], Block);
     output[p--] = Block;
     for (int i = Dn - , j, current_star; i > -; --i)
     {
         current_star = Des[i];
         Vertex[current_star] = true;
         ++Block;
         for (j = ; j < To[current_star].size(); ++j)
             if (Vertex[To[current_star][j]])
                 Union(current_star, To[current_star][j], Block);
         output[p--] = Block;
     }
     for (int i = ; i <= Dn; ++i)
         printf("%d\n", output[i]);
     return ;
 }