POJ 1111 Image Perimeters

http://poj.org/problem?id=1111

 

题意:

给定H行W列由'.'和'X'组成的地图和起点坐标,求与起点八连通的'X'符号组成的图形的总边长

 

解法:

dfs

每到一点计算上下左右'X'的个数得出该点边长,并标记到过的点

 

代码:  724K  0MS

#include <iostream>
#include <cstring>
using namespace std;

#define _ ios_base::sync_with_stdio(0);cin.tie(0)
#define N 21
#define judge(x,y) (x>=0&&x<h&&y>=0&&y<w)

char map[N][N];
bool vis[N][N];
int h, w, sx, sy, ans, mov[8][2] = { { -1, 0}, {0, 1}, {1, 0}, {0, -1}, { -1, -1}, { -1, 1}, {1, 1}, {1, -1}};

void dfs(int x, int y) {
    int xx, yy, cnt = 4;
    vis[x][y] = 1;
    for (int i = 0; i < 4; i++) {
        xx = x + mov[i][0];
        yy = y + mov[i][1];
        if (judge(xx, yy) && map[xx][yy] == 'X')
            cnt--;
    }
    ans += cnt;
    for (int i = 0; i < 8; i++) {
        xx = x + mov[i][0];
        yy = y + mov[i][1];
        if (judge(xx, yy) && !vis[xx][yy] && map[xx][yy] == 'X')
            dfs(xx, yy);
    }
}

int main() {
    _;
    while (cin >> h >> w >> sx >> sy && (h || w || sx || sy)) {
        for (int i = 0; i < h; i++) {
            cin >> map[i];
        }
        ans = 0;
        memset(vis, 0, sizeof(vis));
        dfs(sx - 1, sy - 1);
        cout << ans << endl;
    }
    return 0;
}