hdu-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): 8468    Accepted Submission(s): 2729

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

 

题意：就是Y和M在@相遇的最短的时间。
思路：基本的bfs，先Y搜一次，然后M搜一次，纪录到每一个@的时间，然后求出Y和M在@相遇的最短的时间。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <queue>#define INF 0x3f3f3f3f#define CSH(a, b) memset(a, (b), sizeof(a))#define M 221//数组大小#define N 221//数组大小using namespace std;char s[N][M];int dir[4][2]={-1,0, 1,0, 0,-1, 0,1};int vis1[N][M],vis2[N][M];int m,n;int flag[N][M],flag1[N][M];struct node{    int x;    int y;    int ti;};node fr1,fr2;void intt(){    CSH(vis1,0);    CSH(vis2,0);    for(int i=0;i<n;i++)    {        getchar();        for(int j=0;j<m;j++)        {            scanf("%c",&s[i][j]);            if(s[i][j]=='Y')            {                fr1.x=i;                fr1.y=j;                fr1.ti=0;                vis1[i][j]=1;            }            else if(s[i][j]=='M')            {                fr2.x=i;                fr2.y=j;                fr2.ti=0;                vis2[i][j]=1;            }        }    }}void bfs(){    queue<node> q;    while(!q.empty())    {        q.pop();    }    q.push(fr1);    while(!q.empty())    {        node ne;        ne=q.front();        q.pop();        for(int i=0;i<4;i++)        {            node no;            no.x=ne.x+dir[i][0];            no.y=ne.y+dir[i][1];            no.ti=ne.ti;            if(no.x<n&&no.x>=0&&no.y<m&&no.y>=0&&s[no.x][no.y]!='#'&&!vis1[no.x][no.y])            {                no.ti++;                if(s[no.x][no.y]=='@')                {                    flag[no.x][no.y]=no.ti;                }                vis1[no.x][no.y]=1;                q.push(no);            }        }    }}void bfs1(){    queue<node> q;    while(!q.empty())    {        q.pop();    }    q.push(fr2);    while(!q.empty())    {        node ne;        ne=q.front();        q.pop();        for(int i=0;i<4;i++)        {            node no;            no.x=ne.x+dir[i][0];            no.y=ne.y+dir[i][1];            no.ti=ne.ti;            if(no.x<n&&no.x>=0&&no.y<m&&no.y>=0&&s[no.x][no.y]!='#'&&!vis2[no.x][no.y])//判断边界            {                no.ti++;                if(s[no.x][no.y]=='@')                {                    flag1[no.x][no.y]=no.ti;                }                vis2[no.x][no.y]=1;                //操作..                q.push(no);            }        }    }}int main(){    while(~scanf("%d%d",&n,&m))    {        CSH(flag,-1);        CSH(flag1,-1);        intt();        bfs();        bfs1();        int min=INF;        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            {                if(flag[i][j]+flag1[i][j]<min&&flag[i][j]+flag1[i][j]>0)                {                    min=flag[i][j]+flag1[i][j];                }            }        printf("%d\n",min*11);    }}