hdu4771 Stealing Harry Potter's Precious
注意……你可能会爆内存……
假设一个直接爆搜索词……
队列存储器元件被减少到……
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int inf=(1<<30)-1+(1<<30);
int goal;
char a[110][110];
int dp[110][110][1<<4];
int row,line;
int xd[4]={-1,1,0,0};
int yd[4]={0,0,-1,1};
int state[110][110];
struct node
{
int x,y,s;
node(){}
node(int a,int b,int c){x=a;y=b;s=c;}
};
bool isbeyond(int x,int y)
{
return x<0||x>=row||y<0||y>=line;
}
int bfs(node s)
{
int ans,i;
node now,next;
queue<node>qq;
qq.push(s);
memset(dp,-1,sizeof(dp));
dp[s.x][s.y][s.s]=0;
ans=inf;
while(qq.size())
{
now=qq.front();
qq.pop();
for(i=0;i<4;i++)
{
next=now;
next.x+=xd[i];
next.y+=yd[i];
next.s|=state[next.x][next.y];
if(isbeyond(next.x,next.y))
continue;
if(a[next.x][next.y]=='#')
continue;
if(dp[next.x][next.y][next.s]!=-1)
{
if(dp[now.x][now.y][now.s]+1>=dp[next.x][next.y][next.s])
continue;
}
dp[next.x][next.y][next.s]=dp[now.x][now.y][now.s]+1;
if(next.s==goal)
{
ans=min(ans,dp[next.x][next.y][next.s]);
continue;
}
qq.push(next);
}
}
if(ans==inf)
ans=-1;
return ans;
}
int main()
{
int i,j,n,k;
node s;
while(cin>>row>>line)
{
if(row==0&&line==0)
break;
for(i=0;i<row;i++)
{
cin>>a[i];
for(j=0;j<line;j++)
if(a[i][j]=='@')
s=node(i,j,0);
}
cin>>n;
goal=(1<<n)-1;
memset(state,0,sizeof(state));
for(k=0;k<n;k++)
{
cin>>i>>j;
state[i-1][j-1]|=1<<k;
}
s.s=state[s.x][s.y];
cout<<bfs(s)<<endl;
}
return 0;
}
Stealing Harry Potter's Precious
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1828 Accepted Submission(s): 853
uncle Vernon never allows such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate.
The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:

Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers' properties, so they live in the indestructible rooms and put customers' properties in vulnerable rooms. Harry Potter's precious
are also put in some vulnerable rooms. Dudely wants to steal Harry's things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can't access
the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry's precious are. He wants to collect all Harry's precious by as less steps as possible. Moving from
one room to another adjacent room is called a 'step'. Dudely doesn't want to get out of the bank before he collects all Harry's things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry's precious.
In each test cases:
The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100).
Then a N×M matrix follows. Each element is a letter standing for a room. '#' means a indestructible room, '.' means a vulnerable room, and the only '@' means the vulnerable room from which Dudely starts to move.
The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter's precious in the bank.
In next K lines, each line describes the position of a Harry Potter's precious by two integers X and Y, meaning that there is a precious in room (X,Y).
The input ends with N = 0 and M = 0
2 3
##@
#.#
1
2 2
4 4
#@##
....
####
....
2
2 1
2 4
0 0
-1
5
版权声明:本文博主原创文章。博客,未经同意不得转载。
hdu4771 Stealing Harry Potter's Precious的更多相关文章
- hdu 4771 Stealing Harry Potter's Precious(bfs)
题目链接:hdu 4771 Stealing Harry Potter's Precious 题目大意:在一个N*M的银行里,贼的位置在'@',如今给出n个宝物的位置.如今贼要将全部的宝物拿到手.问最 ...
- hdu 4771 Stealing Harry Potter's Precious
题目:给出一个二维图,以及一个起点,m个中间点,求出从起点出发,到达每一个中间的最小步数. 思路:由于图的大小最大是100*100,所以要使用bfs求出当中每两个点之间的最小距离.然后依据这些步数,建 ...
- hdu4771 Stealing Harry Potter's Precious(DFS,BFS)
练习dfs和bfs的好题. #include<iostream> #include<cstdio> #include<cstdlib> #include<cs ...
- HDU 4771 Stealing Harry Potter's Precious
Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 ...
- HDU 4771 Stealing Harry Potter's Precious (2013杭州赛区1002题,bfs,状态压缩)
Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 ...
- HDU 4771 Stealing Harry Potter's Precious dfs+bfs
Stealing Harry Potter's Precious Problem Description Harry Potter has some precious. For example, hi ...
- hdu 4771 13 杭州 现场 B - Stealing Harry Potter's Precious 暴力bfs 难度:0
Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. W ...
- Stealing Harry Potter's Precious BFS+DFS
Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and hi ...
- 【HDU 4771 Stealing Harry Potter's Precious】BFS+状压
2013杭州区域赛现场赛二水... 类似“胜利大逃亡”的搜索问题,有若干个宝藏分布在不同位置,问从起点遍历过所有k个宝藏的最短时间. 思路就是,从起点出发,搜索到最近的一个宝藏,然后以这个位置为起点, ...
随机推荐
- Java魔法堂:JVM的运行模式 (转)
一.前言 JVM有Client和Server两种运行模式.不同的模式对应不同的应用场景,而JVM也会有相应的优化.本文将记录JVM模式的信息,以便日后查阅. 二.介绍 在$JAVA_HOME/jre/ ...
- osc搜索引擎框架search-framework,TngouDB,gso,
项目目的:OSChina 实现全文搜索的简单封装框架 License: Public Domain 包含内容: 重建索引工具 -> IndexRebuilder.java 增量构建索引工具 -& ...
- curl订单具体解释
为windows假设用户Cygwin模拟unix环境的话,不会有带curl命令,拥有设备,它建议使用Gow为了模拟,它已经自带curl工具,直接安装之后cmd使用环境curl命令可以,由于路径是自己主 ...
- UDP 通信
import java.net.DatagramPacket; import java.net.DatagramSocket; import java.net.InetAddress; public ...
- memwatch的使用
博主的新Blog地址:http://www.brantchen.com 欢迎訪问:) linux下的測试工具真是少之又少,还不好用,近期试用了memwatch,感觉网上的介绍不太好,所以放在这里跟大家 ...
- UVA1450-Airport
题目链接 题意:有一个飞机场.有两条待飞跑到w和e.一条起飞跑道.每一时刻仅仅能起飞一架飞机,然后有w[i]和e[i]架飞机进入w和e跑道.飞机编号从0開始,问说怎样安排起飞能够使得飞机编号的最大值最 ...
- 方案猿身高project联赛,艺术家,相反,养殖场!-------三笔
已经看到了程序猿在电影中都是非常厉害的人物,硬道理键盘噼里啪啦后,奇妙的事情会发生. 当我报了这个专业,開始认真的写程序,在这个领域学习的时候,却发现非常多干这一行 的都自称"码农" ...
- uva11426(莫比乌斯反演)
传送门:GCD Extreme (II) 题意:给定n(n<=4000000),求G G=0 for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) ...
- cocos2d-x CCNode类
文章引用自http://blog.csdn.net/qiurisuixiang/article/details/8763260 1 CCNode是cocos2d-x中一个非常重要的类.CCNode是场 ...
- Android从raw、assets、SD卡中获取资源文件内容
先顺带提一下,raw文件夹中的文件会和project一起经过编译,而assets里面的文件不会~~~ 另外,SD卡获取文件需要权限哦! //从res文件夹中的raw 文件夹中获取文件并读取数据 p ...