N. Find a way

题目网址：http://www.bnuoj.com/v3/contest_show.php?cid=6453#problem/N

题目描述：

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.

Imput:

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 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

Sample Output:

668866

解题思路：两次广搜加枚举到每个KFC店的最短距离，初始化时注意全部为最大，因为存在不可能到达的KFC店，所以如果初始化为0，则枚举时最小距离为0，因此最开始应看成所有的KFC店都不能到达，初始化一个很大的值。就是这点错误，这题WA了4次~~

源代码如下：

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;#define MAXN 200 + 10struct Node {    int x, y;};char MAPT[MAXN][MAXN];int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};int vis[MAXN][MAXN], mdist[MAXN][MAXN], ydist[MAXN][MAXN];int M, N;int check(int xx, int yy) {    int flag = 1;    if (xx < 0 || xx >= M || yy < 0 || yy >= N) flag = 0;    if (vis[xx][yy]) flag = 0;    return flag;}int BFS(int sx, int sy, char ch) {    queue<Node> q;    Node now, pre;    int i, j;    now.x = sx; now.y = sy;    q.push(now);    vis[now.x][now.y] = 1;    while (!q.empty()) {        pre = q.front(); q.pop();        for (i = 0; i < 4; i++) {            now.x = pre.x + dir[i][0];            now.y = pre.y + dir[i][1];            if (check(now.x, now.y) && (MAPT[now.x][now.y] != '#')) {                q.push(now);                vis[now.x][now.y] = 1;                if (ch == 'Y') ydist[now.x][now.y] = ydist[pre.x][pre.y] + 1;                if (ch == 'M') mdist[now.x][now.y] = mdist[pre.x][pre.y] + 1;                //if (MAPT[now.x][now.y] == ch) return dist[now.x][now.y];            }        }    }    return 0;}int main() {    //FILE *p = freopen("test.txt", "r", stdin);    int i, j;    while (scanf("%d%d", &M, &N) != EOF) {        for (i = 0; i < M; i++) {            scanf("%s", MAPT[i]);        }        int sx = 0, sy = 0, ex = 0, ey = 0;        for (i = 0; i < M; i++) {            for (j = 0; j < N; j++) {                if (MAPT[i][j] == 'Y') {                    sx = i; sy = j;                }                if (MAPT[i][j] == 'M') {                    ex = i; ey = j;                }            }        }        memset(vis, 0, sizeof(vis));        memset(ydist, 0, sizeof(ydist));        BFS(sx, sy, 'Y');        memset(vis, 0, sizeof(vis));        memset(mdist, 0, sizeof(mdist));        BFS(ex, ey, 'M');        int ans = 11111111;        for (i = 0; i < M; i++) {            for (j = 0; j < N; j++) {                if (MAPT[i][j] == '@' && mdist[i][j] != 0 && ydist[i][j] != 0) {                    ans = min(ans, mdist[i][j] + ydist[i][j]);                }            }        }        printf("%d\n", ans * 11);    }    return 0;}