Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3451 Accepted Submission(s):
1128
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
Author
能不能用交替进行的方式进行双向广搜？
不行，因为可能你走的可能不是通往最佳的@的。
另加一组数据
6 5
Y.@#.
.#.#.
.#.#.
.#.#.
.#.#.
##M..
77
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 #include<queue> 6 using namespace std; 7 8 int n,m; 9 int yx,yy,mx,my; 10 int to[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; 11 char a[210][210]; 12 int time[2][202][202]; 13 bool hash[2][202][202]; 14 bool flag; 15 struct node 16 { 17 int x,y; 18 int time; 19 }; 20 queue<node>Q[2]; 21 22 int Min(int x,int y) 23 { 24 return x>y? y:x; 25 } 26 bool pd(node &t) 27 { 28 if(t.x>=1&&t.x<=n && t.y>=1&&t.y<=m && a[t.x][t.y]!='#')return false; 29 return true; 30 } 31 int bfs(int x) 32 { 33 int i,hxl=1111111; 34 node t,cur; 35 36 while(!Q[x].empty()) 37 { 38 cur=Q[x].front(); 39 Q[x].pop(); 40 for(i=0;i<4;i++) 41 { 42 t=cur; 43 t.x=t.x+to[i][0]; 44 t.y=t.y+to[i][1]; 45 t.time++; 46 if(pd(t))continue; 47 if(hash[x][t.x][t.y])continue; 48 hash[x][t.x][t.y]=true; 49 time[x][t.x][t.y]=t.time; 50 if(x==1 && a[t.x][t.y]=='@') 51 { 52 hxl=Min(hxl,time[x^1][t.x][t.y]+time[x][t.x][t.y]); 53 } 54 Q[x].push(t); 55 } 56 } 57 return hxl; 58 } 59 void dbfs() 60 { 61 int ans=0; 62 node t; 63 t.x=yx; 64 t.y=yy; 65 t.time=0; 66 Q[0].push(t); 67 hash[0][yx][yy]=true; 68 69 t.x=mx; 70 t.y=my; 71 t.time=0; 72 Q[1].push(t); 73 hash[1][mx][my]=true; 74 75 bfs(0); 76 ans = bfs(1); 77 printf("%d ",ans*11); 78 } 79 int main() 80 { 81 int i,j; 82 while(scanf("%d%d",&n,&m)>0) 83 { 84 for(i=1;i<=n;i++) 85 scanf("%s",a[i]+1); 86 memset(hash,false,sizeof(hash)); 87 memset(time,0,sizeof(time)); 88 for(i=1;i<=n;i++) 89 for(j=1;j<=m;j++){ 90 if(a[i][j]=='Y'){ 91 yx=i; 92 yy=j; 93 } 94 else if(a[i][j]=='M'){ 95 mx=i; 96 my=j; 97 } 98 } 99 while(!Q[0].empty()){ 100 Q[0].pop(); 101 } 102 while(!Q[1].empty()){ 103 Q[1].pop(); 104 } 105 flag=false; 106 dbfs(); 107 } 108 return 0; 109 }