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hihoCoder#1094 Lost in the City

时间:2015-03-21 22:53:22      阅读:179      评论:0      收藏:0      [点我收藏+]

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原题地址

 

限时10s,所以不用考虑什么算法了,暴力吧

分别按照3x3视野的四个方向去地图上匹配,把符合的地点标记出来,最后统一按照从上到下,从左到右的顺序输出。

 

代码:

 1 #include <iostream>
 2 
 3 using namespace std;
 4 
 5 #define MAP_SIZE 250
 6 
 7 bool match(char map[MAP_SIZE][MAP_SIZE], char sight[3][3], int r, int c) {
 8   for (int i = 0; i < 3; i++)
 9     for (int j = 0; j < 3; j++)
10       if (map[r + i][c + j] != sight[i][j])
11         return false;
12   return true;
13 }
14 
15 void find(char map[MAP_SIZE][MAP_SIZE], char sight[3][3], char res[MAP_SIZE][MAP_SIZE], int N, int M) {
16   for (int i = 0; i < N - 2; i++) {
17     for (int j = 0; j < M - 2; j++) {
18       if (match(map, sight, i, j))
19         res[i + 1][j + 1] = 1;
20     }
21   }
22 }
23 
24 void rotate(char sight[3][3]) {
25   for (int i = 0; i < 2; i++)
26     for (int j = 0; j < 2; j++)
27       swap(sight[i][j], sight[2 - j][2 - i]);
28   for (int i = 0; i < 3; i++)
29     swap(sight[0][i], sight[2][i]);
30 }
31 
32 int main() {
33   int N, M;
34   char map[MAP_SIZE][MAP_SIZE];
35   char res[MAP_SIZE][MAP_SIZE];
36   char sight[3][3];
37 
38   cin >> N >> M;
39   for (int i = 0; i < N; i++)
40     for (int j = 0; j < M; j++)
41       cin >> map[i][j];
42 
43   for (int i = 0; i < 3; i++)
44     for (int j = 0; j < 3; j++)
45       cin >> sight[i][j];
46 
47   memset(res, 0, MAP_SIZE * MAP_SIZE * sizeof(char));
48 
49   for (int i = 0; i < 4; i++) {
50     find(map, sight, res, N, M);
51     rotate(sight);
52   }
53 
54   for (int i = 1; i < N - 1; i++)
55     for (int j = 1; j < M - 1; j++)
56       if (res[i][j])
57         cout << i + 1 << " " << j + 1 << endl;
58 
59   return 0;
60 }

 

hihoCoder#1094 Lost in the City

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原文地址:http://www.cnblogs.com/boring09/p/4356398.html

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