Find a way--hdoj
Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4772 Accepted Submission(s): 1624
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.
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
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#
66
88
66#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define min(a,b)(a>b?b:a)
char map[1010][1010];
int vis1[1010][1010],vis2[1010][1010],ans1[1010][1010],ans2[1010][1010];
int m,n;
int dx[4]={1,0,0,-1};
int dy[4]={0,1,-1,0};
struct node
{
int x,y;
int step;
}p,temp;
int judge(node s,int vis[1010][1010])
{
if(s.x<0||s.x>=m||s.y<0||s.y>=n)
return 1;
if(vis[s.x][s.y])
return 1;
if(map[s.x][s.y]=='#')
return 1;
return 0;
}
void dfs(int x,int y,int ans[1010][1010],int vis[1010][1010])
{
p.x=x;
p.y=y;
p.step=0;
vis[x][y]=1;
ans[x][y]=p.step;
queue<node>q;
q.push(p);
while(!q.empty())
{
p=q.front();
q.pop();
for(int i=0;i<4;i++)
{
temp.x=p.x+dx[i];
temp.y=p.y+dy[i];
if(judge(temp,vis))
continue;
temp.step=p.step+1;
ans[temp.x][temp.y]=temp.step;
q.push(temp);
vis[temp.x][temp.y]=1;
}
}
}
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
int i,j,x1,y1,x2,y2;
for(i=0;i<m;i++)
{
scanf("%s",map[i]);
for(j=0;j<n;j++)
{
if(map[i][j]=='Y')
{
x1=i;y1=j;
}
if(map[i][j]=='M')
{
x2=i;y2=j;
}
}
}
memset(vis1,0,sizeof(vis1));
memset(ans1,0,sizeof(ans1));
dfs(x1,y1,ans1,vis1);
memset(vis2,0,sizeof(vis2));
memset(ans2,0,sizeof(ans2));
dfs(x2,y2,ans2,vis2);
int ans=0xfffffff;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
{
if(map[i][j]=='@'&&vis1[i][j]&&vis2[i][j])
{
ans=min(ans,ans1[i][j]+ans2[i][j]);
}
}
printf("%d\n",ans*11);
}
return 0;
}
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