HDU 2612 Find a Way

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612

题目大意：Y，M往@走，问走到最近的@耗时多少。上下左右移动，移动一次11分钟。

解题思路：上来直接枚举@宽搜到Y,M距离记录最小值果断wa，果然搜索还是太耗时了。宽搜两次记录Y到每个@的距离和M到每个@的距离，然后枚举每个点，是@就计算，记录最小值。

这道题最有趣的地方莫过于要脑补限制：

M Y是不能通过的

@是可以通过的

代码：

const int maxn = 500;
const int iadd[] = {0, 1, 0, -1}, jadd[] = {1, 0, -1, 0};
char maze[maxn][maxn];
int vis[maxn][maxn], disy[maxn][maxn], dism[maxn][maxn];
int n, m;
struct node{
    int i, j, t;
};

void bfs1(int sti, int stj){
    memset(vis, 0, sizeof(vis));
    memset(disy, 0x3f, sizeof(disy));
    node u; u.i = sti; u.j = stj; u.t = 0;
    queue<node> q;
    q.push(u);
    vis[u.i][u.j] = 1; disy[u.i][u.j] = 0;
    while(!q.empty()){
        u = q.front(); q.pop();
        for(int i = 0; i < 4; i++){
            node v = u;
            v.i += iadd[i]; v.j += jadd[i]; v.t += 1;
            if(v.i < 0 || v.j < 0 || v.i >= n || v.j >= m) continue;
            if(vis[v.i][v.j]) continue; 
            if(maze[v.i][v.j] == '#') continue;
            if(maze[v.i][v.j] == 'M') continue;
            q.push(v);
            disy[v.i][v.j] = v.t;
            vis[v.i][v.j] = 1;
        }
    }
}

void bfs2(int sti, int stj){
    memset(vis, 0, sizeof(vis));
    memset(dism, 0x3f, sizeof(dism));
    node u; u.i = sti; u.j = stj; u.t = 0;
    queue<node> q;
    q.push(u);
    vis[u.i][u.j] = 1; dism[u.i][u.j] = 0;
    while(!q.empty()){
        u = q.front(); q.pop();
        for(int i = 0; i < 4; i++){
            node v = u;
            v.i += iadd[i]; v.j += jadd[i]; v.t += 1;
            if(v.i < 0 || v.j < 0 || v.i >= n || v.j >= m) continue;
            if(vis[v.i][v.j]) continue; 
            if(maze[v.i][v.j] == '#') continue;
            if(maze[v.i][v.j] == 'Y') continue;
            q.push(v);
            dism[v.i][v.j] = v.t;
            vis[v.i][v.j] = 1;
        }
    }
}

int main(){
    while(scanf("%d %d", &n, &m) != EOF){
        memset(vis, 0, sizeof(vis));
        int yi, yj, mi, mj;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                scanf(" %c", &maze[i][j]);
                if(maze[i][j] == 'Y') yi = i, yj = j;
                if(maze[i][j] == 'M') mi = i, mj = j;
            }
        }
        bfs1(yi, yj);
        bfs2(mi, mj);
        int ans = inf;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(maze[i][j] == '@'){
                    if(disy[i][j] != inf && dism[i][j] != inf)
                        ans = min(ans, disy[i][j] + dism[i][j]);
                }
            }
        }
        printf("%d
", ans * 11);
    }
}

题目：

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16695    Accepted Submission(s): 5358

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