题目链接

Property Distribution Aizu - 0118

思路

dfs

时间复杂度

$$ 每组O(NM) $$

代码

#include <cstdio>
using namespace std;

const int MAXN = 105;

int n, m;
char s[MAXN][MAXN];

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

bool check(int x, int y) {
    if (1 <= x && x <= n && 1 <= y && y <= m) {
        return true;
    } else {
        return false;
    }
}

void dfs(int x, int y, char c) {
    s[x][y] = 'W';
    for (int i = 0; i < 4; i++) {
        int xx = x + dx[i];
        int yy = y + dy[i];
        if (check(xx, yy) && s[xx][yy] == c) {
            dfs(xx, yy, c);
        }
    }
}

int main() {
    while (scanf("%d%d", &n, &m), n + m) {
        for (int i = 1; i <= n; i++) {
            scanf("%s", s[i] + 1);// no &
        }
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (s[i][j] == '#' || s[i][j] == '@' || s[i][j] == '*') {
                    ans++;
                    dfs(i, j, s[i][j]);
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}