hdu5094 Maze
……就是爬管道……
还好内存给的多……
不然就不会做了……
#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<<31)-1;
int dp[51][51][1<<10];
int road[51][51][51][51];
int key[51][51];
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
int n,m,ans;
struct node
{
int x,y,k,step;
node(){}
node(int a,int b,int c,int d){x=a;y=b;k=c;step=d;}
};
bool isin(int x,int y)
{
return x>=1&&x<=n&&y>=1&&y<=m;
}
void bfs()
{
int i;
node now,next;
queue<node>qq;
qq.push(node(1,1,key[1][1],0));
memset(dp,-1,sizeof(dp));
while(qq.size())
{
now=qq.front();
qq.pop();
dp[now.x][now.y][now.k]=now.step;
for(i=0;i<4;i++)
{
next=now;
next.x+=dx[i];
next.y+=dy[i];
next.k|=key[next.x][next.y];
next.step++;
if(!isin(next.x,next.y))
continue;
if(road[now.x][now.y][next.x][next.y]==0)
continue;
if(next.x==n&&next.y==m)
{
ans=min(ans,next.step);
continue;
}
if(road[now.x][now.y][next.x][next.y]>0)
{
if(!((next.k>>road[now.x][now.y][next.x][next.y]-1)&1))
continue;
}
if(dp[next.x][next.y][next.k]!=-1)
{
if(next.step>=dp[next.x][next.y][next.k])
continue;
}
qq.push(next);
}
}
}
int main()
{
int p,sx,sy,ex,ey,t,k,x,y;
while(cin>>n>>m>>p)
{
memset(road,-1,sizeof(road));
cin>>t;
while(t--)
{
cin>>sx>>sy>>ex>>ey>>k;
road[sx][sy][ex][ey]=k;
road[ex][ey][sx][sy]=k;
}
cin>>t;
memset(key,0,sizeof(key));
while(t--)
{
cin>>x>>y>>k;
key[x][y]|=1<<k-1;
}
ans=inf;
bfs();
if(ans==inf)
ans=-1;
cout<<ans<<endl;
}
}
Maze
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 329 Accepted Submission(s): 125
Spock, the deputy captain of Starship Enterprise, fell into Klingon’s trick and was held as prisoner on their mother planet Qo’noS.
The captain of Enterprise, James T. Kirk, had to fly to Qo’noS to rescue his deputy. Fortunately, he stole a map of the maze where Spock was put in exactly.
The maze is a rectangle, which has n rows vertically and m columns horizontally, in another words, that it is divided into n*m locations. An ordered pair (Row No., Column No.) represents a location in the maze. Kirk moves from current location to next costs
1 second. And he is able to move to next location if and only if:
Next location is adjacent to current Kirk’s location on up or down or left or right(4 directions)
Open door is passable, but locked door is not.
Kirk cannot pass a wall
There are p types of doors which are locked by default. A key is only capable of opening the same type of doors. Kirk has to get the key before opening corresponding doors, which wastes little time.
Initial location of Kirk was (1, 1) while Spock was on location of (n, m). Your task is to help Kirk find Spock as soon as possible.
Each test case consists of several lines. Three integers are in the first line, which represent n, m and p respectively (1<= n, m <=50, 0<= p <=10).
Only one integer k is listed in the second line, means the sum number of gates and walls, (0<= k <=500).
There are 5 integers in the following k lines, represents xi1, yi1, xi2, yi2, gi; when gi >=1, represents there is a gate of type gi between location (xi1, yi1) and (xi2,
yi2); when gi = 0, represents there is a wall between location (xi1, yi1) and (xi2, yi2), ( | xi1 - xi2 | + | yi1 - yi2 |=1, 0<= gi <=p
)
Following line is an integer S, represent the total number of keys in maze. (0<= S <=50).
There are three integers in the following S lines, represents xi1, yi1 and qi respectively. That means the key type of qi locates on location (xi1, yi1), (1<= qi<=p).
If there is no possible plan, output -1.
4 4 9
9
1 2 1 3 2
1 2 2 2 0
2 1 2 2 0
2 1 3 1 0
2 3 3 3 0
2 4 3 4 1
3 2 3 3 0
3 3 4 3 0
4 3 4 4 0
2
2 1 2
4 2 1
14
hdu5094 Maze的更多相关文章
- Backtracking algorithm: rat in maze
Sept. 10, 2015 Study again the back tracking algorithm using recursive solution, rat in maze, a clas ...
- (期望)A Dangerous Maze(Light OJ 1027)
http://www.lightoj.com/volume_showproblem.php?problem=1027 You are in a maze; seeing n doors in fron ...
- 1204. Maze Traversal
1204. Maze Traversal A common problem in artificial intelligence is negotiation of a maze. A maze ...
- uva705--slash maze
/*这道题我原本是将斜线迷宫扩大为原来的两倍,但是在这种情况下对于在斜的方向上的搜索会变的较容易出错,所以参考了别人的思路后将迷宫扩展为原来的3倍,这样就变成一般的迷宫问题了*/ #include&q ...
- HDU 4048 Zhuge Liang's Stone Sentinel Maze
Zhuge Liang's Stone Sentinel Maze Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/327 ...
- Borg Maze(MST & bfs)
Borg Maze Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9220 Accepted: 3087 Descrip ...
- poj 3026 bfs+prim Borg Maze
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9718 Accepted: 3263 Description The B ...
- HDU 4035:Maze(概率DP)
http://acm.split.hdu.edu.cn/showproblem.php?pid=4035 Maze Special Judge Problem Description When w ...
- POJ 3026 : Borg Maze(BFS + Prim)
http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS Memory Limit: 65536K Total Submissions ...
随机推荐
- tensorflow的卷积和池化层(二):记实践之cifar10
在tensorflow中的卷积和池化层(一)和各种卷积类型Convolution这两篇博客中,主要讲解了卷积神经网络的核心层,同时也结合当下流行的Caffe和tf框架做了介绍,本篇博客将接着tenso ...
- __class__属性与元类
class M(type): def __str__(self): return "gege" aa = "ccf" cc = "ccc" ...
- Servlet笔记9--转发与重定向
关于Web中资源跳转的问题: 转发和重定向基本代码: package com.bjpowernode.javaweb; import java.io.IOException; import javax ...
- Submatrix Sum
Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return ...
- Java多态概述
多态 所谓多态,实际上就是一个对象的多种状态: 下面例子中,Tiger可以看做Tiger,也可以看做Animal Cat 可以看做Cat,也可以看做Animal Dog 可以看做Dog,也可以看做A ...
- SNMP中MIB2所有主要节点
系统组:system组包含以下对象集(.1.3.6.1.2.1.1): 对象名:sysDescr(1) OID:system.1 对象类型:DisplayString[255] 访问模式:只读 描述: ...
- 【PAT】1032 Sharing (25)(25 分)
1032 Sharing (25)(25 分) To store English words, one method is to use linked lists and store a word l ...
- 【LOJ】#2109. 「JLOI2015」骗我呢
题解 我深思熟虑许久才算是明白个大概的计数问题吧 先是转化成一个矩形,列一条直线y = x,y = x - (m + 1)我们从(0,0)走到(n + m + 1,m + 1)就是答案 因为我们起始相 ...
- 003.KVM虚拟机部署-CentOS6.8
一 实验说明 CentOS 7的KVM虚拟机推荐使用qcow2磁盘格式,本实验在KVM中安装CentOS 6.8 64虚拟机. 二 命令部署过程 2.1 上传镜像 使用winscp上传至/data/i ...
- 003.SMB相关文件
一 常用文件 /etc/samba/smb.conf #主配置文件 /etc/samba/lmhosts #对应NetBIOS名与主机IP的文件,samba会自动搜索(只对本机生效) /etc/sam ...