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codeforces 510B. Fox And Two Dots 解题报告

时间:2015-02-16 11:31:25      阅读:182      评论:0      收藏:0      [点我收藏+]

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题目链接:http://codeforces.com/problemset/problem/510/B

题目意思:给出 n 行 m 列只有大写字母组成的字符串。问具有相同字母的能否组成一个环。

  很容易知道要用到深搜。暴力搜索~~~

  

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 const int maxn = 50 + 5;
 8 char g[maxn][maxn];
 9 bool vis[maxn][maxn];
10 
11 int dx[] = {0, 1, 0, -1};
12 int dy[] = {1, 0, -1, 0};
13 
14 int n, m;
15 
16 bool dfs(int x, int y, int px, int py, char c)
17 {
18     vis[x][y] = 1;
19     for (int i = 0; i < 4; i++)    {
20         int tx = x + dx[i];
21         int ty = y + dy[i];
22         if (tx == px && ty == py)    // 和上一次走过的点冲突
23             continue;
24         if (tx >= 0 && tx < n && ty >= 0 && ty < m && g[tx][ty] == c) {
25             if (vis[tx][ty])     // 形成环
26                 return 1;
27             if (dfs(tx, ty, x, y, c))
28                 return 1;
29         }
30     }
31     return 0;
32 }
33 
34 
35 int main()
36 {
37     while (scanf("%d%d", &n, &m) != EOF) {
38         for (int i = 0; i < n; i++)
39             scanf("%s", g[i]);
40         memset(vis, 0, sizeof(vis));
41 
42         bool flag = false;
43         for (int i = 0; i < n && !flag; i++) {
44             for (int j = 0; j < m && !flag; j++) {
45                 if (!vis[i][j]) {
46                     if (dfs(i, j, -1, -1, g[i][j])) {
47                         flag = true;
48                         break;
49                     }
50                 }
51             }
52         }
53         printf("%s\n", flag ? "Yes" : "No");
54     }
55     return 0;
56 }

 

cgy4ever 的代码:

http://ideone.com/udz3bN

技术分享
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n, m;
 5 string board[51];
 6 bool visited[51][51];
 7 bool findCycle = false;
 8 int dx[] = {1, -1, 0, 0};
 9 int dy[] = {0, 0, 1, -1};
10 
11 void dfs(int x, int y, int fromX, int fromY, char needColor)
12 {
13     if(x < 0 || x >= n || y < 0 || y >= m) return;
14     if(board[x][y] != needColor) return;
15     if(visited[x][y])
16     {
17         findCycle = true;
18         return;
19     }
20     visited[x][y] = true;
21     for(int f = 0; f < 4; f++)
22     {
23         int nextX = x + dx[f];
24         int nextY = y + dy[f];
25         if(nextX == fromX && nextY == fromY) continue;
26         dfs(nextX, nextY, x, y, needColor);
27     }
28 }
29 
30 int MAIN()
31 {
32     cin >> n >> m;
33     for(int i = 0; i < n; i++)
34         cin >> board[i];
35     memset(visited, false, sizeof(visited));
36     for(int i = 0; i < n; i++)
37         for(int j = 0; j < m; j++)
38             if(!visited[i][j])
39                 dfs(i, j, -1, -1, board[i][j]);
40     cout << (findCycle ? "Yes" : "No") << endl;
41     return 0;
42 }
43 
44 int main()
45 {
46     #ifdef LOCAL_TEST
47         freopen("in.txt", "r", stdin);
48         freopen("out.txt", "w", stdout);
49     #endif
50     ios :: sync_with_stdio(false);
51     cout << fixed << setprecision(16);
52     return MAIN();
53 }
View Code

 

codeforces 510B. Fox And Two Dots 解题报告

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原文地址:http://www.cnblogs.com/windysai/p/4293748.html

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