Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E: Rock, Scissors, Paper

Bart's sister Lisa has created a new civilization on a two-dimensional grid. At the outset each grid location may be occupied by one of three life forms: Rocks, Scissors, or Papers. Each day, differing life forms occupying horizontally or vertically adjacent grid locations wage war. In each war, Rocks always defeat Scissors, Scissors always defeat Papers, and Papers always defeat Rocks. At the end of the day, the victor expands its territory to include the loser's grid position. The loser vacates the position.

Your job is to determine the territory occupied by each life form after n days. The first line of input contains t, the number of test cases. Each test case begins with three integers not greater than 100: r and c, the number of rows and columns in the grid, and n. The grid is represented by the r lines that follow, each with c characters. Each character in the grid is R, S, or P, indicating that it is occupied by Rocks, Scissors, or Papers respectively.

For each test case, print the grid as it appears at the end of the nth day. Leave an empty line between the output for successive test cases.

Sample Input

2
3 3 1
RRR
RSR
RRR
3 4 2
RSPR
SPRS
PRSP

Output for Sample Input

RRR
RRR
RRR

RRRS
RRSP
RSPR

---

題目描述：
一個盤面上有剪刀石頭布，每過一個時間後，每個格子會跟鄰近的四個格子對決，
贏者將會佔領那隔，問 d 時間後的盤面。

題目解法：

類似生命遊戲，看每一格下一秒會變成什麼樣子，看鄰近有沒有贏過他的，如果有就替換。

#include <stdio.h>
#include <algorithm>
using namespace std;
char g[2][1024][1024];
int main() {
    int testcase;
    int n, m, d;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d", &n, &m, &d);
        while(getchar() != '\n');
        int i, j, k, flag = 0;
        for(i = 0; i < n; i++)
            gets(g[0][i]);
        while(d--) {
            for(i = 0; i < n; i++) {
                for(j = 0; j < m; j++) {
                    g[!flag][i][j] = g[flag][i][j];
                    if(g[flag][i][j] == 'R') {
                        if(i+1 < n && g[flag][i+1][j] == 'P')
                            g[!flag][i][j] = 'P';
                        else if(i-1 >= 0 && g[flag][i-1][j] == 'P')
                            g[!flag][i][j] = 'P';
                        else if(j+1 < m && g[flag][i][j+1] == 'P')
                            g[!flag][i][j] = 'P';
                        else if(j-1 >= 0 && g[flag][i][j-1] == 'P')
                            g[!flag][i][j] = 'P';
                    }
                    else if (g[flag][i][j] == 'P') {
                        if(i+1 < n && g[flag][i+1][j] == 'S')
                            g[!flag][i][j] = 'S';
                        else if(i-1 >= 0 && g[flag][i-1][j] == 'S')
                            g[!flag][i][j] = 'S';
                        else if(j+1 < m && g[flag][i][j+1] == 'S')
                            g[!flag][i][j] = 'S';
                        else if(j-1 >= 0 && g[flag][i][j-1] == 'S')
                            g[!flag][i][j] = 'S';
                    }
                    else {
                        if(i+1 < n && g[flag][i+1][j] == 'R')
                            g[!flag][i][j] = 'R';
                        else if(i-1 >= 0 && g[flag][i-1][j] == 'R')
                            g[!flag][i][j] = 'R';
                        else if(j+1 < m && g[flag][i][j+1] == 'R')
                            g[!flag][i][j] = 'R';
                        else if(j-1 >= 0 && g[flag][i][j-1] == 'R')
                            g[!flag][i][j] = 'R';
                    }
                }
            }
            flag = !flag;
        }
        for(i = 0; i < n; i++)
            g[flag][i][m] = '\0', puts(g[flag][i]);
        if(testcase)    puts("");
    }
    return 0;
}