模拟->YY POJ 2339 Rock, Scissors, Paper

Rock, Scissors, PaperTime Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ2339DescriptionBart'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 horizontallyor 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 loservacates the position.Your job is to determine the territory occupied by each life form after n days.InputThe 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 thatfollow, 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.OutputFor 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

Sample Output

RRR
RRR
RRR

RRRS
RRSP
RSPR

#include <iostream>#include <stack>#include <string>#include <string.h>#include <queue>#include <cmath>#include <map>#include <cstdio>using namespace std;char mapto[101][101];char maptoo[101][101];int main(){int t, r, c, n;cin >> t;while (cin >> r >> c >> n){for (int i = 0; i < r; i++){for (int j = 0; j < c; j++){cin >> mapto[i][j];maptoo[i][j] = mapto[i][j];}}if (n == 0){for (int i = 0; i < r; i++){for (int j = 0; j < c; i++){cout << mapto[i][j];}cout << endl;}}else{for (int k = 0; k < n; k++){for (int i = 0; i < r; i++){for (int j = 0; j < c; j++){if (mapto[i][j] == 'S'){if (i + 1 < r){if (mapto[i + 1][j] == 'R'){maptoo[i][j] = 'R';}if (mapto[i + 1][j] == 'P'){maptoo[i + 1][j] = 'S';}}if (i - 1 >= 0){if (mapto[i - 1][j] == 'R'){maptoo[i][j] = 'R';}if (mapto[i - 1][j] == 'P'){maptoo[i - 1][j] = 'S';}}if (j + 1 < c){if (mapto[i][j + 1] == 'R'){maptoo[i][j] = 'R';}if (mapto[i][j + 1] == 'P'){maptoo[i][j + 1] = 'S';}}if (j - 1 >= 0){if (mapto[i][j - 1] == 'R'){maptoo[i][j] = 'R';}if (mapto[i][j - 1] == 'P'){maptoo[i][j - 1] = 'S';}}}else if (mapto[i][j] == 'R'){if (i + 1 < r){if (mapto[i + 1][j] == 'S'){maptoo[i + 1][j] = 'R';}if (mapto[i + 1][j] == 'P'){maptoo[i][j] = 'P';}}if (i - 1 >= 0){if (mapto[i - 1][j] == 'S'){maptoo[i - 1][j] = 'R';}if (mapto[i - 1][j] == 'P'){maptoo[i][j] = 'P';}}if (j + 1 < c){if (mapto[i][j + 1] == 'S'){maptoo[i][j + 1] = 'R';}if (mapto[i][j + 1] == 'P'){maptoo[i][j] = 'P';}}if (j - 1 >= 0){if (mapto[i][j - 1] == 'S'){maptoo[i][j - 1] = 'R';}if (mapto[i][j - 1] == 'P'){maptoo[i][j] = 'P';}}}else{if (i + 1 < r){if (mapto[i + 1][j] == 'S'){maptoo[i][j] = 'S';}if (mapto[i + 1][j] == 'R'){maptoo[i + 1][j] = 'P';}}if (i - 1 >= 0){if (mapto[i - 1][j] == 'S'){maptoo[i][j] = 'S';}if (mapto[i - 1][j] == 'R'){maptoo[i - 1][j] = 'P';}}if (j + 1 < c){if (mapto[i][j + 1] == 'S'){maptoo[i][j] = 'S';}if (mapto[i][j + 1] == 'R'){maptoo[i][j + 1] = 'P';}}if (j - 1 >= 0){if (mapto[i][j - 1] == 'S'){maptoo[i][j] = 'S';}if (mapto[i][j - 1] == 'R'){maptoo[i][j - 1] = 'P';}}}}}for (int i = 0; i < r; i++){for (int j = 0; j < c; j++){mapto[i][j] = maptoo[i][j];}}}for (int i = 0; i < r; i++){for (int j = 0; j < c; j++){cout << mapto[i][j];}cout << endl;}cout << endl;}}return 0;}

思路：剪头石子布,按个模拟,顺序征服。

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