hdu 2612 多终点BFS

Find a way

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

 

思路：两次BFS存下对每个kfc的最短距离，之后两两相加取min

代码：

#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define mj
#define db double
#define ll long long
using namespace std;
const int N=1e8+2;
const int mod=1e9+7;
//const ll inf=1e16+10;
#define inf 0x3f3f3f
typedef pair<int,int> P;
int n,m;
char  s[300][300];
int d[300][300],k[205][205];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
int t[2][205*205];
int bfs(int sx,int sy,int id)
{
    queue<P> q;
    for(int i=0;i<205;i++){
        for(int j=0;j<205;j++){
            d[i][j]=N;
        }
    }
    q.push(P(sx,sy));
    d[sx][sy]=0;
    while(q.size()){
        P p;
        p=q.front(),q.pop();
        for(int i=0;i<4;i++){
            int nx=p.first+dx[i],ny=p.second+dy[i];
            if(0<=nx&&nx<n&&0<=ny&&ny<m&&s[nx][ny]!='#'&&d[nx][ny]==N){
                d[nx][ny]=d[p.first][p.second]+1;
                if(s[nx][ny]=='@') t[id][k[nx][ny]]=d[nx][ny];//到该点的距离
                q.push(P(nx,ny));
            }
        }
    }
    return 0;
}
int main()
{
    int xx[3],yy[3];
    while(scanf("%d%d",&n,&m)==2){
        memset(t,inf, sizeof(t));
        int c=0,cnt=0,ma=N;
        for(int i=0;i<n;i++){
            scanf("%s",s[i]);
            for(int j=0;j<m;j++){
                if(s[i][j]=='Y') xx[c]=i,yy[c++]=j;
                else if(s[i][j]=='M') xx[c]=i,yy[c++]=j;
                else if(s[i][j]=='@'){
                    k[i][j]=cnt++;
                }
            }
        }
        bfs(xx[0],yy[0],0),bfs(xx[1],yy[1],1);
        for(int i=0;i<cnt;i++){
            ma=min(t[0][i]+t[1][i],ma);
//            printf("%d %d
",t[0][i],t[1][i]);
        }
        printf("%d
",11*ma);
    }
    return 0;
}