Hdu 2612 Find a way(bfs) Find a way

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概述

Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4044 Accepted Submission(s): 1348

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

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题意：

求M和Y到达同一个‘@’所需要的时间总和最小，每步需要时间11分钟。

题解：

两次BFS求M和Y到每个‘@’之间的最小距离，之后求相加的最小值。

CODE：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos(-1.0);
typedef long long ll;
const int INF=1000010;
using namespace std;
struct node
{
int x,y;
int num;//保存两人到@的最小值
int num2;
};
vector<node>q;
queue<node>que;
int n,m;
char mp[222][222];
bool vis[222][222];
int len,Min,l;
int Mx,My,Yx,Yy;
int xx[4]= {-1,0,1,0};
int yy[4]= {0,1,0,-1};
void bfs(int x,int y)
{
memset(vis,0,sizeof vis);
while(que.size())
que.pop();
node t,tt;
t.x=x,t.y=y;
t.num=0;
que.push(t);
int s=0;
while(que.size())
{
t=que.front();
que.pop();
if(mp[t.x][t.y]=='@')
{
for(int i=0; i<l; i++)//找到@
{
if(q[i].x==t.x&&q[i].y==t.y)
{
q[i].num=min(t.num,q[i].num);
break;
}
}
}
for(int i=0; i<4; i++)
{
tt.x=t.x+xx[i];
tt.y=t.y+yy[i];
tt.num=t.num+1;
if(tt.x>=0&&tt.x<n&&tt.y>=0&&tt.y<m&&mp[tt.x][tt.y]!='#'&&!vis[tt.x][tt.y])
{
que.push(tt);
vis[tt.x][tt.y]=1;
}
}
}
}
int main()
{
//freopen("test.in","r",stdin);
while(cin>>n>>m)
{
len=0;
q.clear();
node a;
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
{
cin>>mp[i][j];
if(mp[i][j]=='@')
{
a.x=i,a.y=j,a.num=INF,a.num2=0;
q.push_back(a);
}
if(mp[i][j]=='Y')
{
Yx=i,Yy=j;
}
if(mp[i][j]=='M')
{
Mx=i,My=j;
}
}
l=q.size();
bfs(Mx,My);
for(int i=0; i<l; i++)//把num的值转给num2，num重新复制无穷大
{
q[i].num2=q[i].num;
q[i].num=INF;
}
bfs(Yx,Yy);
Min=INF;
for(int i=0; i<l; i++)
{
if(q[i].num2+q[i].num<Min)
Min=q[i].num+q[i].num2;
}
cout<<Min*11<<endl;
}
return 0;
}