201512-2 消除类游戏

实现

#include <cstdio>

#define MAXN 32

int chess[MAXN][MAXN];
int marks[MAXN][MAXN];

int main() {
    int row, col;

    scanf("%d%d",&row,&col);

    for (int i = 0;i < row;++i) {
        for (int j = 0;j < col;++j) {
            scanf("%d",&chess[i][j]);
        }
    }
    
    for (int i = 0;i < row;++i) {
        int pre_val = 0;
        int same_cnt = 0;
        for (int j = 0;j < col;++j) {
            if (pre_val == chess[i][j]) {
                same_cnt+=1;
                if (j == col - 1 && same_cnt >= 2) {
                    for (;same_cnt >= 0; --same_cnt) {
                        marks[i][j - same_cnt] = 1;
                    }
                }
            } else if (pre_val != chess[i][j]){
                if(same_cnt >= 2) {
                    for (;same_cnt >= 0; --same_cnt) {
                        marks[i][j - same_cnt - 1] = 1;
                    }
                }
                same_cnt = 0;
            } 
            pre_val = chess[i][j];
        }
    }

    for (int j = 0;j < col;++j) {
        int pre_val = 0;
        int same_cnt = 0;
        for (int i = 0;i < row;++i) {
            if (pre_val == chess[i][j]) {
                same_cnt+=1;
                if ( i == row - 1 && same_cnt >= 2) {
                    for (;same_cnt >= 0; --same_cnt) {
                        marks[i - same_cnt][j] = 1;
                    }
                }
            }  else if (pre_val != chess[i][j]) {
                if (same_cnt >= 2) {
                    for (;same_cnt >= 0;--same_cnt) {
                        marks[i - same_cnt - 1][j] = 1;
                    }
                }
                same_cnt = 0;
            }
            pre_val = chess[i][j];
        }
    }

    for (int i = 0;i < row;++i) {
        for (int j = 0;j < col;++j) {
            if (marks[i][j] == 1) {
                printf("0 ");
            } else {
                printf("%d ",chess[i][j]);
            }
        }
        printf("\n");
    }

}
posted @ 2020-08-30 13:32  amonqsq  阅读(115)  评论(0编辑  收藏  举报