[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 = 0; 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 = 0;
    scanf("%d %d", &V, &E);
    for (int i = 0, 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 = 0; i < Dn; Vertex[Des[i]] = false, ++i)
        scanf("%d", &Des[i]);
    Block = V - Dn;
    for (int i = 0, j; i < V; ++i)
        if (Vertex[i])
            for (j = 0; j < To[i].size(); ++j)
                if (Vertex[To[i][j]])
                    Union(i, To[i][j], Block);
    output[p--] = Block;
    for (int i = Dn - 1, j, current_star; i > -1; --i)
    {
        current_star = Des[i];
        Vertex[current_star] = true;
        ++Block;
        for (j = 0; 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 = 0; i <= Dn; ++i)
        printf("%d
", output[i]);
    return 0;
}