hdu 1312 Red and Black

//简单的的搜索,找到能扩充的最大步数
#include<iostream>
using namespace std;

const int maxn = 20 + 5;
int dirx[4] = {0, 0, -1, 1};
int diry[4] = {-1, 1, 0, 0};
int n, m, ans;
int sx, sy;
char map[maxn][maxn];
bool used[maxn][maxn];

int dfs(int x, int y){
int px, py;
int max = 0;
used[x][y] = 1;
ans++;
for(int i=0; i<4; i++){
px = dirx[i] + x;
py = diry[i] + y;
if(!used[px][py] && map[px][py] != '#' && px >= 1 && px <= m && py >= 1 && py <= n){
dfs(px, py);
}
}
return ans;
}

int main(){
while(scanf("%d%d", &n, &m) != EOF){
if(n == 0 && m == 0)
break;
memset(map, '#', sizeof(map));
ans = 0;
getchar();
for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
scanf("%c", &map[i][j]);
if(map[i][j] == '@'){
sx = i;
sy = j;
}
}
getchar();
}
memset(used, 0, sizeof(used));
ans = 0;
int sum = dfs(sx, sy);
cout<<sum<<endl;
}
return 0;
}

转载于:https://www.cnblogs.com/cbyniypeu/p/3518087.html