题目链接

Cheese Aizu - 0558

思路

基础题，多次bfs。

时间复杂度

$$ O(NMK) $$

代码

#include <cstdio>
#include <iostream>
#include <queue>

using namespace std;

const int MAXN = 1010;

char s[MAXN][MAXN];
int step[MAXN][MAXN];
int n, m, k;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

pair<int, int> find(char c) {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (s[i][j] == c) {
                return make_pair(i, j);
            }
        }
    }
    return make_pair(0, 0);
}

void init() {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            step[i][j] = 0;
        }
    }
}

bool check(int x, int y) {
    if (1 <= x && x <= n && 1 <= y && y <= m && s[x][y] != 'X') {
        return true;
    } else {
        return false;
    }
}

int bfs(pair<int, int> cur, char c) {
    init();
    queue<pair<int, int> > que;
    que.push(cur);
    while(!que.empty()) {
        cur = que.front();
        que.pop();
        for (int i = 0; i < 4; i++) {
            int xx = cur.first + dx[i];
            int yy = cur.second + dy[i];
            if (check(xx, yy) && step[xx][yy] == 0) {
                step[xx][yy] = step[cur.first][cur.second] + 1;
                que.push(make_pair(xx, yy));
                if (s[xx][yy] == c) {
                    return step[xx][yy];
                }
            }
        }
    }
    return 0;
}

int main() {

    scanf("%d%d%d", &n, &m, &k);// don't forget &
    for (int i = 1; i <= n; i++) {
        scanf("%s", s[i] + 1);// no &
    }

    pair<int, int> cur = find('S');
    int ans = 0;
    for (int i = 1; i <= k; i++) {
        ans += bfs(cur, '0' + i);
        cur = find('0' + i);
    }
    printf("%d\n", ans);
    return 0;
}