hdu 4775 Infinite Go(暴力)

题目链接：hdu 4775 Infinite Go

题目大意：两个人下围棋，总共走了n步。黑棋和白棋交替走，假设一片棋的上下左右被封死，那么该片棋子就会被吃掉，问说最后黑白棋各剩多少个。

解题思路：比較恶心的模拟题，相邻同样色的棋子要用并查集连接。而且要记录每片棋子还剩的空格数。假设空格数为0的话说明该片棋子被其它颜色围住，则要剔除掉，不且将相邻的位置不同色的棋空格数加1。主要是细节上的问题。

例子
8
7
5 5
4 5
3 5
3 4
4 4
3 3
4 6
18
1 3
1 4
2 2
1 5
2 4
2 3
3 1
3 2
3 5
3 4
4 2
4 3
4 4
1 6
5 3
3 3
1 10
3 3
12
1 2
1 1
2 1
2 2
1 3
3 1
2 3
1 4
3 2
3 3
4 2
2 4
4
1 1
1 2
2 2
2 1
4
2000000000 2000000000
2000000000 1999999999
1999999999 1999999999
1999999999 2000000000
8
1 2
4 1
2 1
4 2
2 3
4 3
3 2
2 2
17
1 3
1 4
2 2
1 5
2 4
2 3
3 1
3 2
3 5
3 4
4 2
4 3
4 4
1 6
5 3
30 30
3 3
17
1 3
1 4
2 2
1 5
2 4
2 3
3 1
3 2
3 5
3 4
4 2
3 3
4 4
1 6
5 3
4 3
100 100
答案
4 2
9 4
6 4
1 2
2 2
4 3
9 4
9 3

#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 1e4;
const int INF = 2*1e9+10;
const int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
typedef pair<int, int> pii;

int N, Nw, Nb, X[maxn+5], Y[maxn+5], f[maxn+5], c[maxn+5];
map<pii, int> R;

void init () {
    scanf("%d", &N);

    Nw = N / 2;
    Nb = N - Nw;

    R.clear();
    for (int i = 0; i < N; i++) {
        f[i] = i;
        c[i] = 0;
    }
}

inline int bit (int x) {
    return x&1;
}

int getfar (int x) {
    return f[x] == x ? x : f[x] = getfar(f[x]);
}

inline bool isEmpty (int x, int y) {
    if (x <= 0 || y <= 0 || x >= INF || y >= INF)
        return false;
    if (R.count(make_pair(x, y)))
        return false;
    return true;
}

inline int count_empty (pii u) {
    int cnt = 0;
    for (int i = 0; i < 4; i++) {
        int x = u.first + dir[i][0];
        int y = u.second + dir[i][1];
        if (isEmpty(x, y))
            cnt++;
    }
    return cnt;
}

inline void link_board (int x, int y) {
    int fx = getfar(x);
    int fy = getfar(y);

    f[fy] = fx;
    c[fx] += c[fy];
    /**/
    c[fx]--;
}

int del_board (int col, int x, int y) {
    int cnt = 1;
    pii u = make_pair(x, y);
    queue<pii> que;
    que.push(u);

    f[R[u]] = R[u];
    R.erase(u);

    while (!que.empty()) {
        u = que.front();
        que.pop();

        for (int i = 0; i < 4; i++) {
            int p = u.first + dir[i][0];
            int q = u.second + dir[i][1];

            if (p <= 0 || p >= INF || q <= 0 || q >= INF)
                continue;

            pii v = make_pair(p, q);
            if (!R.count(v))
                continue;

            int set = R[v];

            if (bit(set) != col) {
                int k = getfar(set);
                c[k]++;
                continue;
            }

            f[R[v]] = R[v];
            R.erase(v);
            cnt++;
            que.push(v);
        }
    }
    return cnt;
}

 void del_empty (int k) {
    int fk = getfar(k);
    c[fk]--;

    if (c[fk] == 0) {
        int set = bit(fk);
        int cnt = del_board(set, X[fk], Y[fk]);
        if (set)
            Nw -= cnt;
        else
            Nb -= cnt;
    }
}

void solve () {

    for (int i = 0; i < N; i++) {
        scanf("%d%d", &X[i], &Y[i]);
        pii v = make_pair(X[i], Y[i]);
        c[i] = count_empty(v);
        R[v] = i;

        for (int j = 0; j < 4; j++) {
            int p = X[i] + dir[j][0];
            int q = Y[i] + dir[j][1];

            if (p <= 0 || q <= 0 || p >= INF || q >= INF)
                continue;

            pii u = make_pair(p, q);
            if (!R.count(u))
                continue;

            int k = R[u];
            if (bit(i) == bit(k))
                link_board(i, k);
            else
                del_empty(k);
        }

        int fi = getfar(i);
        if (c[fi] == 0) {
            int cnt = del_board(bit(fi), X[fi], Y[fi]);
            if (bit(fi))
                Nw -= cnt;
            else
                Nb -= cnt;
        }
    }
    printf("%d %d\n", Nb, Nw);
}

int main () {
    int cas;    
    scanf("%d", &cas);
    while (cas--) {
        init();
        solve();
    }
    return 0;
}