HDU2612-Find a way

Find a way

                                                                              Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                     Total Submission(s): 12710    Accepted Submission(s): 4077

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

 

Sample Output

668866

 

Author

yifenfei

 

Source

奋斗的年代

 

题意：两个人到KFC见面，两个人在不同的位置，KFC也不只一个，对每个人BFS广搜，求出每个人到各个KFC的最短距离，最后对于多个KFC地点，求出两个人到这的最小步数的和，取最小值就可以了。

#include <iostream>#include <queue>#include <stdio.h>#include <string.h>using namespace std;#define N 210const int INF=0x3f3f3f3f;int m,n,mark[N][N],dis[N][N][2],flag;int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};char s[N][N];struct node{    int x,y,step;};bool judge(int x,int y){    if(x>=0&&x<m&&y>=0&&y<n&&s[x][y]!='#'&&mark[x][y]==0)        return 1;    return 0;}void bfs(int x,int y){    queue<node>q;    node cur,next;    cur.x=x;cur.y=y;    cur.step=0;    mark[x][y]=1;    q.push(cur);    while(!q.empty())    {        cur=q.front();        q.pop();        next.step=cur.step+1;        for(int k=0; k<4; k++)        {            next.x=x=cur.x+dir[k][0];            next.y=y=cur.y+dir[k][1];            if(judge(x,y))            {                mark[x][y]=1;                if(s[x][y]=='@')                    dis[x][y][flag]=next.step;                q.push(next);            }        }    }}int main(){    int mi;    while(~scanf("%d %d",&m,&n))    {        mi=INF;        for(int i=0; i<m; i++)            for(int j=0; j<n; j++)                dis[i][j][0]=dis[i][j][1]=INF;        for(int i=0; i<m; i++)            scanf("%s",s[i]);        for(int i=0; i<m; i++)            for(int j=0; j<n; j++)            {                if(s[i][j]=='Y')                {                    flag=0;                    memset(mark,0,sizeof(mark));                    bfs(i,j);                }                else if(s[i][j]=='M')                {                    flag=1;                    memset(mark,0,sizeof(mark));                    bfs(i,j);                }            }        for(int i=0; i<m; i++)            for(int j=0; j<n; j++)                if(s[i][j]=='@'&&mi>dis[i][j][0]+dis[i][j][1])                    mi=dis[i][j][0]+dis[i][j][1];        printf("%d\n",mi*11);    }    return 0;}