Poj 3026 Borg Maze
2016-01-16 15:33
351 查看
Borg Maze
Description
The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated
subspace network that insures each member is given constant supervision and guidance.
Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is
that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost
of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.
Input
On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character,
a space `` '' stands for an open space, a hash mark ``#'' stands for an obstructing wall, the capital letter ``A'' stand for an alien, and the capital letter ``S'' stands for the start of the search. The perimeter of the maze is always closed, i.e., there
is no way to get out from the coordinate of the ``S''. At most 100 aliens are present in the maze, and everyone is reachable.
Output
For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.
Sample Input
Sample Output
Source
Svenskt Mästerskap i
Programmering/Norgesmesterskapet 2001
思路:先用BFS将各个点的最小距离算出来,然后再求最小生成树
AC代码如下:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11235 | Accepted: 3691 |
The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated
subspace network that insures each member is given constant supervision and guidance.
Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is
that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost
of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.
Input
On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character,
a space `` '' stands for an open space, a hash mark ``#'' stands for an obstructing wall, the capital letter ``A'' stand for an alien, and the capital letter ``S'' stands for the start of the search. The perimeter of the maze is always closed, i.e., there
is no way to get out from the coordinate of the ``S''. At most 100 aliens are present in the maze, and everyone is reachable.
Output
For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.
Sample Input
2 6 5 ##### #A#A## # # A# #S ## ##### 7 7 ##### #AAA### # A# # S ### # # #AAA### #####
Sample Output
8 11
Source
Svenskt Mästerskap i
Programmering/Norgesmesterskapet 2001
思路:先用BFS将各个点的最小距离算出来,然后再求最小生成树
AC代码如下:
#include <iostream> #include <cstring> #include <cmath> #include <queue> #include <cstdio> #define INF 0x3f3f3f3f using namespace std; const int maxn=100+5; char xs[maxn][maxn]; int mp[maxn][maxn]; int node[maxn][maxn]; int dis[maxn],vis1[maxn],vis2[maxn][maxn]; int xx[]={1,-1,0,0}; int yy[]={0,0,1,-1}; struct Node{ int x,y; int n; }tmp,now; void bfs(int x,int y,int n,int m){ queue<Node> Q; memset(vis2,0,sizeof(vis2)); while(!Q.empty()) Q.pop(); tmp.x=x;tmp.y=y;tmp.n=0; Q.push(tmp); vis2[x][y]=1; while(!Q.empty()){ now=Q.front();Q.pop(); if(node[now.x][now.y] && (now.x!=x || now.y!=y)){ mp[node[x][y]][node[now.x][now.y]]=now.n; } for(int i=0;i<4;i++){ int tx=now.x+xx[i],ty=now.y+yy[i]; if(tx<0 || tx>m || ty<0 || ty>n || vis2[tx][ty]) continue; if(xs[tx][ty]!='#'){ Node ttmp; ttmp.x=tx,ttmp.y=ty; vis2[tx][ty]=1; ttmp.n=now.n+1; Q.push(ttmp); } } } } void prim(int n){ int ans=0; memset(dis,INF,sizeof(dis)); memset(vis1,0,sizeof(vis1)); for(int i=1;i<=n;i++){ dis[i]=mp[1][i]; } vis1[1]=1,dis[1]=0; for(int i=1;i<n;i++){ int minn=INF,p=0; for(int j=1;j<=n;j++){ if(!vis1[j] && dis[j]<minn){ minn=dis[j]; p=j; } } ans+=minn;vis1[p]=1; for(int j=1;j<=n;j++){ if(!vis1[j] && dis[j]>mp[p][j]){ dis[j]=mp[p][j]; } } } cout<<ans<<endl; } int main(){ int t; char tt[20]; cin>>t; while(t--){ int n,m; cin>>n>>m; gets(tt); for(int i=0;i<m;i++){ gets(xs[i]); } int num=0; memset(node,0,sizeof(node)); for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ if(xs[i][j]=='S' || xs[i][j]=='A'){ node[i][j]=++num; } } } memset(mp,INF,sizeof(mp)); for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ if(node[i][j]) bfs(i,j,n,m); } } prim(num); } return 0; }
相关文章推荐
- C++子类如何调用父类构造函数
- ag(The silver search)的使用
- uboot编译学习---执行make TQ2440_config后的,操作过程
- Makefile万能写法(gcc程序以及arm-linux-gcc程序)-转
- 2016 最佳 Linux 发行版排行榜
- [Leetcode]23. Merge k Sorted Lists @python
- 浅析gcc、arm-linux-gcc和arm-elf-gcc的关系-转
- java实现的简单网页爬虫:Servlet + MySQL5.5(二)
- UBOOT编译学习-----配置过程
- python读取写入文件
- JS魔法堂:彻底理解0.1 + 0.2 === 0.30000000000000004的背后
- 【转】iOS 文件下载及断点续传
- JavaScript: 2015 年回顾与展望
- 左右点击分页方法
- git branch的使用
- iOS UITableViewCell 展开实现
- Unity中的世界坐标和NGUI中的坐标的相互转换
- Xcode生命周期简介
- 快速排序
- linux添加字体的过程