poj2339- 简单模拟
来源:互联网 发布:php is numeric绕过 编辑:程序博客网 时间:2024/06/06 02:10
感觉自己的代码不够简洁啊,以后要提炼。。。
#include <stdio.h>#include <stdlib.h>#include <string.h>char rsp[2][110][110];int r, s, p;void update(int k, int i, int j){if (i - 1 >= 0){if (rsp[(k+1)%2][i][j] == 'R' && rsp[(k+1)%2][i - 1][j] == 'P'){rsp[k][i][j] = 'P';}if (rsp[(k+1)%2][i][j] == 'P' && rsp[(k+1)%2][i - 1][j] == 'S'){rsp[k][i][j] = 'S';}if (rsp[(k+1)%2][i][j] == 'S' && rsp[(k+1)%2][i - 1][j] == 'R'){rsp[k][i][j] = 'R';}}if (j - 1 >= 0){if (rsp[(k+1)%2][i][j] == 'R' && rsp[(k+1)%2][i][j - 1] == 'P'){rsp[k][i][j] = 'P';}if (rsp[(k+1)%2][i][j] == 'P' && rsp[(k+1)%2][i][j - 1] == 'S'){rsp[k][i][j] = 'S';}if (rsp[(k+1)%2][i][j] == 'S' && rsp[(k+1)%2][i][j - 1] == 'R'){rsp[k][i][j] = 'R';}}if (i + 1 <= r){if (rsp[(k+1)%2][i][j] == 'R' && rsp[(k+1)%2][i + 1][j] == 'P'){rsp[k][i][j] = 'P';}if (rsp[(k+1)%2][i][j] == 'P' && rsp[(k+1)%2][i + 1][j] == 'S'){rsp[k][i][j] = 'S';}if (rsp[(k+1)%2][i][j] == 'S' && rsp[(k+1)%2][i + 1][j] == 'R'){rsp[k][i][j] = 'R';}}if (j + 1 <= s){if (rsp[(k+1)%2][i][j] == 'R' && rsp[(k+1)%2][i][j + 1] == 'P'){rsp[k][i][j] = 'P';}if (rsp[(k+1)%2][i][j] == 'P' && rsp[(k+1)%2][i][j + 1] == 'S'){rsp[k][i][j] = 'S';}if (rsp[(k+1)%2][i][j] == 'S' && rsp[(k+1)%2][i][j + 1] == 'R'){rsp[k][i][j] = 'R';}}}int main(){int t;scanf("%d", &t);while (t --){int k = 1;memset(rsp, 0 , sizeof(rsp));scanf("%d %d %d", &r, &s, &p);for (int i = 0; i < r; ++ i){scanf("%s", rsp[0][i]);}for (int l = 0; l < p; ++ l){for (int i = 0; i < r; ++ i){for (int j = 0; j < s; ++ j){rsp[k][i][j] = rsp[(k+1)%2][i][j];update(k, i, j);}}k = (k + 1) % 2;}k = (k + 1) % 2;for (int i = 0; i < r; ++ i){printf("%s\n", rsp[k][i]);}if (t != 0){printf("\n");}}return 0;}
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