POJ 1027 The Same Game 大暴消

题目中给出了优先级 当四连块的size相同时 优先消左下角的

具体做题思路很简单

1、找出优先级最大的size最大的连通块 这一步显然是Floodfill求

2、比较鬼畜的输出答案...

3、消除连通块 同样可以Floodfill来求

4、更新整幅图 这也是本题最恶心的地方 一个特别繁琐的模拟...

Tips：鉴于本题输出实在太鬼畜...

          我在这题上因为各种输出上的坑点WA了一下午...

          不过我也费了好大劲搞到了官方的测试数据...这里共享给大家...

          http://download.youkuaiyun.com/detail/sci_m3/9681676

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};

char map[20][20];
bool vis[20][20];
int x, y, tmp; char c;

inline void remove(int x, int y) {
    char c = map[x][y];
    map[x][y] = '.';
    for(int i = 0; i < 4; ++i) {
        int xx = x + dx[i], yy = y + dy[i];
        if(xx >= 0 && yy >= 0 && xx < 10 && yy < 15 && map[xx][yy] == c)
            remove(xx, yy);
    }
}

inline void floodfill(int x, int y) {
    tmp++; vis[x][y] = true;
    for(int i = 0; i < 4; ++i) {
        int xx = x + dx[i], yy = y + dy[i];
        if(xx >= 0 && yy >= 0 && xx < 10 && yy < 15 && !vis[xx][yy] && map[xx][yy] == map[x][y])
            floodfill(xx, yy);
    }
}

inline int find(int &x, int &y, char &c) {
    int size = 0; memset(vis, false, sizeof(vis));
    for(int j = 0; j < 15; ++j) for(int i = 0; i < 10; ++i) {
        if(!vis[i][j] && map[i][j] != '.') {
            tmp = 0; floodfill(i, j);
            if(tmp > size) { size = tmp; x = i; y = j; c = map[i][j]; }
        }
    }
    return size;
}

inline void f5() {
    for(int j = 0; j < 15; ++j) {
    	int tmp = 0;
    	for(int i = 0; i < 10; ++i) if(map[i][j] == '.') tmp++;
        for(int i = 0; i < 10 - tmp; ++i) while (map[i][j] == '.') {
        	int k = i;
            while (k != 9) { swap(map[k][j], map[k + 1][j]); k++; }
        }
    }
    bool change[20]; int tmpx = 0;
    memset(change, false, sizeof(change));
    for (int j = 0; j < 15; ++j) {
        int tmp = 0;
        for(int i = 0; i < 10; ++i) if(map[i][j] == '.') tmp++;
        if (tmp == 10) { change[j] = true; tmpx++; }
    }
    for (int j = 0; j < 15 - tmpx; ++j) while (change[j]) {
        int k = j;
        while (k != 14) {
            for(int i = 0; i < 10; ++i) swap(map[i][k], map[i][k + 1]);
            swap(change[k],change[k + 1]); k++;
        }
    }
}

int main() {
    int T; scanf("%d",&T);
    for(int cas = 1; cas <= T; ++cas) {
        for (int i = 9; i >= 0; --i) scanf("%s", map[i]);
        printf("Game %d:\n\n", cas);
        int leave = 150, step = 0, score = 0;
        for(;;) {
            int ans = find(x, y, c);
            if(ans <= 1) break;
            printf("Move %d at (%d,%d): removed %d balls of color %c, got %d points.\n", ++step, x + 1, y + 1, ans, c, (ans - 2) * (ans - 2));
            leave -= ans;
            score += (ans - 2) * (ans - 2);
            remove(x, y);
            f5();
        }
        printf("Final score: %d, with %d balls remaining.\n\n",leave == 0 ? score + 1000 : score, leave);
    }
    return 0;
}