HDU Problem 2612 Find a way 【BFS】

Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10324 Accepted Submission(s): 3365

Problem Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

Sample Input

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

Author

yifenfei

Source

奋斗的年代

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就因为一个字母的，没注意，WA了超十次。一开始我用“@”找Y，M，比较每次的最小值，结果超时。然后换战略，用Y，M找“@”，只用广搜两次，每次把搜到的结果储存一下。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

struct node{
    int x, y, step;
}a, b, temp;
bool vis[300][300];
char maze[300][300];
int dx[4] = {0,1,0,-1};
int dy[4] = {1,0,-1,0};
int ans[300][300], ans1[300][300];
int m, n, temp1, temp2, sx1, sx2, sy1, sy2;
int bfs(int x, int y, bool flag) {
    queue<node> que;
    memset(vis, false, sizeof(vis));
    while (!que.empty()) que.pop();
    temp.x = x; temp.step = 0;
    temp.y = y;
    vis[x][y] = true;
    que.push(temp);
    while (que.size()) {
        a = que.front();
        que.pop();
        if (maze[a.x][a.y] == '@') {
            //找到后就记录结果。
            if (flag) ans1[a.x][a.y] = a.step;
            else ans[a.x][a.y] = a.step;
        }
        for (int i = 0; i < 4; i++) {
            b.x = a.x + dx[i];
            b.y = a.y + dy[i];
            b.step = a.step + 1;
            if (b.x < 0 || b.y < 0 || b.x >= n || b.y >= m || maze[b.x][b.y] == '#')
                continue;
            if (vis[b.x][b.y]) continue;
            vis[b.x][b.y] = true;
            que.push(b);
        }
    }
}
int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        memset(ans, 0, sizeof(ans));
        memset(ans1, 0, sizeof(ans1));
        for (int i = 0; i < n; i++) {
            scanf("%s", maze[i]);
            for (int j = 0; j < m; j++) {
                if (maze[i][j] == 'Y') {
                    sx1 = i, sy1 = j;
                }
                else if (maze[i][j] == 'M') {
                    sx2 = i, sy2 = j;
                }
            }
        }
        bool flag = false;
        bfs(sx1, sy1, flag);
        bfs(sx2, sy2, flag = true);
        int cnt = 1e12;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                //必须这个点被同时访问到
                if(ans1[i][j] != 0 && ans[i][j] != 0)
                    cnt = min(cnt, ans[i][j] + ans1[i][j]);
            }
        }
        printf("%d
", cnt*11);
    }
    return 0;
}