POJ - 2339 Rock, Scissors, Paper
初看题目时就发了个错误，我因为没有耐心看题而不了解题目本身的意思，找不到做题的突破口，即使看了一些题解，还是没有想到方法。
后来在去问安叔，安叔一语道破天机，问我有没有搞清题目的意思，我才恍然大悟，做题当然要先搞懂题目意思，不然做什么题呢！以后一定要再三注意！而且要把每一个细节都注意到！
POJ - 2339 Rock, Scissors, Paper
Time Limit: 5000MS |
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Memory Limit: 65536KB |
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64bit IO Format: %I64d & %I64u |
Description
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.
Input
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.
Output
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
Sample Output
RRR
RRR
RRR
RRRS
RRSP
RSPR
Source
1 #include<stdio.h> 2 #include<string.h> 3 #include<math.h> 4 #include<stdlib.h> 5 char toda[105][105]; 6 char tomo[105][105]; 7 int main() 8 { 9 int t = 0, r = 0, c = 0, n = 0, i = 0, j = 0; 10 scanf("%d", &t); 11 while(t--) { 12 scanf("%d%d%d", &r, &c, &n); 13 for(i = 0; i < r; i++) { 14 scanf("%s", toda[i]); 15 } 16 for(i = 0; i < r; i++) { 17 for(j = 0; j < c; j++) { 18 tomo[i][j] = toda[i][j]; 19 } 20 } 21 while(n--) { 22 for(i = 0; i < r; i++) { 23 for(j = 0; j < c; j++) { 24 if(toda[i][j] == 'R') { 25 if(j+1 < c && toda[i][j+1] == 'S') 26 tomo[i][j+1] = 'R'; 27 if(i-1 >= 0 && toda[i-1][j] == 'S') 28 tomo[i-1][j] = 'R'; 29 if(j-1 >= 0 && toda[i][j-1] == 'S') 30 tomo[i][j-1] = 'R'; 31 if(i+1 < r && toda[i+1][j] == 'S') 32 tomo[i+1][j] = 'R'; 33 } 34 else if(toda[i][j] == 'S') { 35 if(j+1 < c && toda[i][j+1] == 'P') 36 tomo[i][j+1] = 'S'; 37 if(i-1 >= 0 && toda[i-1][j] == 'P') 38 tomo[i-1][j] = 'S'; 39 if(j-1 >= 0 && toda[i][j-1] == 'P') 40 tomo[i][j-1] = 'S'; 41 if(i+1 < r && toda[i+1][j] == 'P') 42 tomo[i+1][j] = 'S'; 43 } 44 else { 45 if(j+1 < c && toda[i][j+1] == 'R') 46 tomo[i][j+1] = 'P'; 47 if(i-1 >= 0 && toda[i-1][j] == 'R') 48 tomo[i-1][j] = 'P'; 49 if(j-1 >= 0 && toda[i][j-1] == 'R') 50 tomo[i][j-1] = 'P'; 51 if(i+1 < r && toda[i+1][j] == 'R') 52 tomo[i+1][j] = 'P'; 53 } 54 } 55 } 56 for(i = 0; i < r; i++) { 57 for(j = 0; j < c; j++) { 58 toda[i][j] = tomo[i][j]; 59 } 60 } 61 } 62 for(i = 0; i < r; i++) { 63 for(j = 0; j < c; j++) { 64 printf("%c", toda[i][j]); 65 } 66 printf("\n"); 67 } 68 if(n) printf("\n"); 69 } 70 return 0; 71 }
