HDU 2612 - Find a way - [BFS]

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=2612

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

 

題意：

兩人分別從 "Y" 和 "M" 出發，在地圖上走，要找一個 "@" 碰頭，每走一格花費時間 $11$ 分鐘，求最短的碰頭時間。

 

題解：

以兩我的分別爲起點作BFS，求出兩人到每家KFC的各自花費的時間。

對所有KFC維護兩人到達時間之和的最小值即爲答案。

 

AC代碼：

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
const int MAX=205;
const int dx[4]={0,1,0,-1};
const int dy[4]={1,0,-1,0};
int n,m;
char mp[MAX][MAX];
pii Y,M;
int KFCcnt;
map<pii,int> KFC;
int Time[2][MAX*MAX];

queue<pii> Q;
int d[MAX][MAX];
void bfs(pii st,int t)
{
    memset(Time[t],0x3f,sizeof(Time[t]));
    memset(d,-1,sizeof(d));
    Q.push(st); d[st.first][st.second]=0;
    while(!Q.empty())
    {
        pii now=Q.front(); Q.pop();
        if(mp[now.first][now.second]=='@')
            Time[t][KFC[now]]=d[now.first][now.second];
        for(int k=0;k<4;k++)
        {
            pii nxt=now; nxt.first+=dx[k], nxt.second+=dy[k];
            if(mp[nxt.first][nxt.second]=='#') continue;
            if(~d[nxt.first][nxt.second]) continue;
            Q.push(nxt);
            d[nxt.first][nxt.second]=d[now.first][now.second]+11;
        }
    }
}
int main()
{
    for(int i=0;i<MAX;i++) mp[i][0]=mp[0][i]='#';
    while(cin>>n>>m)
    {
        KFCcnt=0; KFC.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%s",mp[i]+1), mp[i][m+1]='#';
            for(int j=1;j<=m;j++)
            {
                if(mp[i][j]=='@') KFC[make_pair(i,j)]=++KFCcnt;
                if(mp[i][j]=='Y') Y=make_pair(i,j);
                if(mp[i][j]=='M') M=make_pair(i,j);
            }
        }
        for(int j=0;j<=m+1;j++) mp[n+1][j]='#';

        bfs(Y,0);
        bfs(M,1);
        int res=0x3f3f3f3f;
        for(int i=1;i<=KFCcnt;i++) res=min(res,Time[0][i]+Time[1][i]);
        printf("%d\n",res);
    }
}