hdu 2612
来源:互联网 发布:淘宝同款排除王工具 编辑:程序博客网 时间:2024/05/20 22:39
Find a way
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12932 Accepted Submission(s): 4141
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
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12932 Accepted Submission(s): 4141
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
#include<cstdio>#include<iostream>#include<cstring>#include<queue>using namespace std;int dir[4][2]={1,0, -1,0, 0,1, 0,-1};int ori[211][211];int map[211][211];int x_s1,y_s1;int x_s2,y_s2;int n,m;int ans[211][211];struct node{int x,y;int step;};int min(int a,int b){return a>b?b:a;}int judge(int x,int y){if(x<0 || x>=n || y<0 || y>=m)return 1;if(map[x][y]==1)return 1;return 0;}void BFS(int x,int y){queue<node>q;node cur,next;int i;cur.x=x;cur.y=y;cur.step=0;map[cur.x][cur.y]=1;q.push(cur);while(!q.empty()){cur=q.front();q.pop();if(ori[cur.x][cur.y]==2)ans[cur.x][cur.y]+=cur.step;for(i=0;i<4;i++){next.x=cur.x+dir[i][0];next.y=cur.y+dir[i][1];if(judge(next.x,next.y))continue;next.step=cur.step+1;map[next.x][next.y]=1;q.push(next);}}}int main(){int i,l;char str[211];int Ans;while(scanf("%d%d",&n,&m)!=-1){memset(ori,0,sizeof(ori));memset(ans,0,sizeof(ans));for(i=0;i<n;i++){scanf("%s",str);for(l=0;str[l];l++){if(str[l]=='Y'){x_s1=i;y_s1=l;}else if(str[l]=='M'){x_s2=i;y_s2=l;}else if(str[l]=='.')ori[i][l]=0;else if(str[l]=='#')ori[i][l]=1;else if(str[l]=='@')ori[i][l]=2;}}for(i=0;i<n;i++)for(l=0;l<m;l++)map[i][l]=ori[i][l];BFS(x_s1,y_s1);for(i=0;i<n;i++)for(l=0;l<m;l++)map[i][l]=ori[i][l];BFS(x_s2,y_s2);Ans=11111111;for(i=0;i<n;i++){for(l=0;l<m;l++){if(ans[i][l]){Ans=min(Ans,ans[i][l]);}}}printf("%d\n",Ans*11);}return 0;}
0 0
- HDU 2612
- hdu 2612
- hdu 2612
- HDU 2612
- HDU 2612
- HDU 2612
- HDU 2612
- hdu 2612
- hdu 2612
- HDU 2612
- HDU 2612 两次bfs
- HDU-2612(双BFS)
- HDU 2612 两次BFS
- HDU 2612 BFS*2
- hdu 2612 Beat
- hdu 2612(dfs)
- hdu~2612(bfs)
- hdu-2612 两次bfs
- (7)IDEA 14.1 安装MyBatis插件
- Xcode Cannot create __weak reference in file using manual reference counting
- C#中Timer的使用
- 【BZOJ2301】【HAOI2011】Problem b 莫比乌斯反演
- OFBiz 数据库自增SequenceUtil.java
- hdu 2612
- Android.mk的用法和基础 && m、mm、mmm编译命令
- 关于Xml文件读取报错的问题
- Configure ISSUE LIST
- CentOS安转C/C++环境
- error: org.springframework.web.util.WebAppRootLi
- 关于Oracle SQL中系统时间和库里数据时间的比较
- Linux下套接字详解(四)----简单的TCP套接字应用(迭代型)
- opencv频域相乘