标签：des style blog color os io strong for
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 22409 | Accepted: 12100 |
Description
Input
Output
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
一道简单的DFS题，统计可以搜索的个数。题意：一男子站在一个黑色的瓷砖上，他旁边有黑色和红色的瓷砖，但他只能走黑色的瓷砖，而且是4个方向，问他最多能走多少个黑色的瓷砖。
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; int num; char map[25][25]; int visited[25][25]; int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0}; void DFS(int x, int y, int n, int m) { int mx, my; for(int i = 0; i<4; i++) { mx = x+dir[i][0]; my = y+dir[i][1]; if(mx>=1 && mx<=n && my>=1 && my<=m && !visited[mx][my] && map[mx][my]==‘.‘) { visited[mx][my] = 1; num++; DFS(mx, my, n, m); } } } int main() { int n, m, x, y; while(scanf("%d%d", &m, &n)!=EOF && (n || m)) { num = 1; memset(visited, 0, sizeof(visited)); for(int i = 1; i<=n; i++) { for(int j = 1; j<=m; j++) { cin>>map[i][j]; if(map[i][j] == ‘@‘) { x = i; y = j; visited[x][y] = 1; } } } DFS(x, y, n, m); printf("%d\n", num); } return 0; }
poj 1979 && zoj 2165 Red and Black,布布扣,bubuko.com
poj 1979 && zoj 2165 Red and Black
标签：des style blog color os io strong for
原文地址：http://www.cnblogs.com/fengxmx/p/3910876.html
