POJ 1027 The Same Game 大暴消

来源：互联网发布：优秀程序员简历模板编辑：程序博客网时间：2024/06/07 02:26

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.csdn.net/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;}

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