hdoj 2612 Find a way【bfs+队列】
Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5401 Accepted Submission(s):
1823
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.
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
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
两个起点，多个终点，两个人必须到达同一个终点，开两个数组，分别存放第一个和第二个人到达终点的步数，然后在对应终点出让两人的步数相加，取最小的
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 210
#define INF 0x3f3f3f
using namespace std;
int bu1[MAX][MAX];//记录第一个人步数
int bu2[MAX][MAX];//记录第二个人步数
int p;
char map[MAX][MAX];
int vis[MAX][MAX];
int n,m;
struct node
{
int x,y;
int step;
};
int MIN(int x,int y)
{
return x<y?x:y;
}
void bfs(int x1,int y1,int p)
{
memset(vis,0,sizeof(vis));
int j,i,ok=0;
int move[4][2]={0,1,0,-1,1,0,-1,0};
queue<node>q;
node beg,end;
beg.x=x1;
beg.y=y1;
beg.step=0;
q.push(beg);
while(!q.empty())
{
end=q.front();
q.pop();
if(map[end.x][end.y]=='@')//遇见@则表示到达终点
{
if(p==1)
bu1[end.x][end.y]=end.step;
else
bu2[end.x][end.y]=end.step;
}
for(i=0;i<4;i++)
{
beg.x=end.x+move[i][0];
beg.y=end.y+move[i][1];
if(!vis[beg.x][beg.y]&&0<=beg.x&&beg.x<n&&0<=beg.y&&beg.y<m&&map[beg.x][beg.y]!='#')
{
vis[beg.x][beg.y]=1;
beg.step=end.step+11;
q.push(beg);
}
}
}
}
int main()
{
int sum,j,i,t,k,x1,x2,y1,y2,min;
int s[11000];
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
}
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(map[i][j]=='Y')
{
x1=i;y1=j;
}
else if(map[i][j]=='M')
{
x2=i;y2=j;
}
}
}
memset(bu1,INF,sizeof(bu1));
bfs(x1,y1,1);
memset(bu2,INF,sizeof(bu2));
bfs(x2,y2,2);
min=INF;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(bu1[i][j]!=INF&&bu2[i][j]!=INF)
{
min=MIN(bu1[i][j]+bu2[i][j],min);//取两者步数和的最小值
}
}
}
printf("%d\n",min);
}
return 0;
}
