poj 1979 Red and Black -- dfs

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Description

Input

Output

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

题意是从@点出发，然后遍历整个图，‘.’表示可以到达的地方，求最多可以走多少个‘.’，初始的点也算一个，注意sum=1初始化

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char map[25][25];
int vis[25][25];
int n,m,num;
int dr[4]={0,1,-1,0};
int dc[4]={-1,0,0,1};
void dfs(int x,int y){
	int i;
	for(i=0;i<4;i++){
		int dx=x+dr[i];
		int dy=y+dc[i];
		if(dx>=0&&dx<n&&dy>=0&&dy<m&&!vis[dx][dy]&&map[dx][dy]=='.'){
			vis[dx][dy]=1;
			num++;
			dfs(dx,dy);
		}
	}
}
int main(){
	int i,j,x,y;
	//freopen("test.txt","r",stdin);
	while(~scanf("%d %d",&m,&n)){
		if(m==0&&n==0)
			break;
		char c[25];
		for(i=0;i<n;i++){
			scanf("%s",&c);
			for(j=0;j<m;j++){
				map[i][j]=c[j];
				if(map[i][j]=='@'){
					x=i;y=j;
				}
			}
		}
		//printf("%d %d\n",x,y);
		memset(vis,0,sizeof(vis));
		num=1;
		dfs(x,y);
		printf("%d\n",num);
	}
	return 0;
}

poj 1979 Red and Black -- dfs

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原文地址：http://blog.csdn.net/qq_27717967/article/details/51365344