hdu--2576
2015-11-01 12:10
295 查看
Description
If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find
the girl, then you can date with the girl.Else the girl will date with other boys. What a pity!
The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there.
There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move
left, right, up or down.
Input
The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).
The next r line is the map’s description.
Output
For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".
Sample Input
Sample Output
解题思路:和迷宫题类似,题目大意为从Y出发是否能走到G的位置,如果可以输出最少时间,否则输出Please give me another chance!,但注意的时,当走到石头的地方时间为k的倍数时,是可以通过的,否则不能通过。对于Bfs题目最难的就是什么时候该标记的问题,这里只有当走过的地方为K的倍数时,才标记。
提供一组测试数据:
2 4 5
..Y#
.##G
6
代码如下:
If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find
the girl, then you can date with the girl.Else the girl will date with other boys. What a pity!
The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there.
There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move
left, right, up or down.
Input
The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).
The next r line is the map’s description.
Output
For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".
Sample Input
1 6 6 2 ...Y.. ...#.. .#.... ...#.. ...#.. ..#G#.
Sample Output
7
解题思路:和迷宫题类似,题目大意为从Y出发是否能走到G的位置,如果可以输出最少时间,否则输出Please give me another chance!,但注意的时,当走到石头的地方时间为k的倍数时,是可以通过的,否则不能通过。对于Bfs题目最难的就是什么时候该标记的问题,这里只有当走过的地方为K的倍数时,才标记。
提供一组测试数据:
2 4 5
..Y#
.##G
6
代码如下:
#include<stdio.h> #include<queue> #include<string.h> #define INF 0x3f3f3f3f using namespace std; int n,m,k; char mark[110][110]; int vist[110][110]; int dis[4][2]={{-1,0},{1,0},{0,1},{0,-1}}; int ans; struct stu { int x,y,t; }; stu s,e; void bfs() { memset(vist,0,sizeof(vist)); //vist[s.x][s.y]=1; stu temp; stu vt; queue<stu>q; while(!q.empty()) { q.pop(); } q.push(s); while(!q.empty()) { vt=q.front(); q.pop(); for(int i=0;i<4;i++) { temp.x=vt.x+dis[i][0]; temp.y=vt.y+dis[i][1]; temp.t=vt.t+1; if(temp.x>=0&&temp.x<n&&temp.y>=0&&temp.y<m) { if(!vist[temp.x][temp.y]) { if(mark[temp.x][temp.y]!='#'||temp.t%k==0) { if(temp.x==e.x&&temp.y==e.y) { ans=temp.t; return; } else { if(temp.t%k==0) vist[temp.x][temp.y]=1; q.push(temp); } } } } } } } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&m,&k); for(int i=0;i<n;i++) { getchar(); for(int j=0;j<m;j++) { scanf("%c",&mark[i][j]); if(mark[i][j]=='Y') { s.x=i; s.y=j; s.t=0; } else if(mark[i][j]=='G') { e.x=i; e.y=j; } } } if(s.x==e.x&&s.y==e.y) { printf("0\n"); continue; } ans=INF; bfs(); if(ans>=INF) printf("Please give me another chance!\n"); else printf("%d\n",ans); } return 0; }
相关文章推荐
- OC 学习笔记3 文件夹下所有.txt文件的行数
- hdu 5523 Game 【BestCoder Round #61 (div.2)】
- 英文标点
- Android开发者网址导航
- JAVA基础2-标识符、关键字、数据类型与运算符
- STL 中 lower_bound 与 upper_bound 与 二分查找
- uva 10570——Meeting with Aliens
- android studio中api 23的消息推送机制
- [leetcode 246] Strobogrammatic Number
- Android团队
- 【Linux 内核】文件系统(概念篇)
- 【疯狂的架构】牛公司组织结构图一览:华为,阿里,腾讯,百度,新浪……
- Codeforces Round #328 (Div. 2) B. The Monster and the Squirrel
- HDU--3466(0-1背包+贪心/后效性)
- 【鸟哥的linux私房菜-学习笔记】Bash shell之管线命令
- ThinkPHP3.2.3中典型的ajax获取json数据方法
- codeforces #328 C. The Big Race
- 如何做到一个分数序列和
- python os目录和文件相关操作
- Coursera_Stanford_ML_ex7_K-means and PCA 作业记录