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
- HDU 1501 Zipper 字符串
题目大意:输入有一个T,表示有T组测试数据,然后输入三个字符串,问第三个字符串能否由第一个和第二个字符串拼接而来,拼接的规则是第一个和第二个字符串在新的字符串中的前后的相对的顺序不能改变,问第三个字符 ...
- 第11月第3天 直播 rtmp yuv
1. LiveVideoCoreSDK AudioUnitRender ==> MicSource::inputCallback ==> GenericAudioMixer::pushBu ...
- 如何得到Slave应用relay-log的时间
官方社区版MySQL 5.7.19 基于Row+Position搭建的一主一从异步复制结构:Master->{Slave} ROLE HOSTNAME BASEDIR DATADIR IP PO ...
- mysql-8.0.11-winx64 免安装版配置方法
mysql-8.0.11-winx64.zip 下载地址:https://dev.mysql.com/downloads/file/?id=476233 mysql-8.0.11-winx64.zi ...
- 数据库SQL中case when函数的用法
Case具有两种格式,简单Case函数和Case搜索函数.这两种方式,可以实现相同的功能.简单Case函数的写法相对比较简洁,但是和Case搜索函数相比,功能方面会有些限制,比如写判断式. 简单Cas ...
- tomcat启动报错:Injection of autowired dependencies failed
tomcat启动报错:Injectjion of autowired dependencies failed 环境: 操作系统:centos6.5 tomcat: 7.0.52 jdk:openjdk ...
- Go 2 Draft Designs
Go 2 Draft Designs 28 August 2018 Yesterday, at our annual Go contributor summit, attendees got a sn ...
- 【linux】监控磁盘情况并自动删除备份文件
背景:我有一个备份目录/home/kzy/bakup,会每天备份一些信息.随着日子一天天的过去,这个文件夹越来越大,终于把磁盘撑满了..... 需求:当磁盘占有率超过80%时自动删除该文件夹下最老的3 ...
- TcxGrid Sqlite text类型 显示memo
- Spring对象依赖关系处理
Spring中给对象属性赋值 1.通过set方法给属性注入值 2.p名称空间 3.自动装配 4.注解 编写MVCModel调用userAction MVCModel public class MVCM ...