hdu 1044(bfs+状压)
非常经典的一类题型
没有多个出口。这里题目没有说清楚
Collect More Jewels
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4684 Accepted Submission(s): 983
Your goddess The Lady seeks to possess the Amulet, and with it to gain deserved ascendance over the other gods.
You, a newly trained Rambler, have been heralded from birth as the instrument of The Lady. You are destined to recover the Amulet for your deity, or die in the attempt. Your hour of destiny has come. For the sake of us all: Go bravely with The Lady!
If you have ever played the computer game NETHACK, you must be familiar with the quotes above. If you have never heard of it, do not worry. You will learn it (and love it) soon.
In this problem, you, the adventurer, are in a dangerous dungeon. You are informed that the dungeon is going to collapse. You must find the exit stairs within given time. However, you do not want to leave the dungeon empty handed. There are lots of rare jewels in the dungeon. Try collecting some of them before you leave. Some of the jewels are cheaper and some are more expensive. So you will try your best to maximize your collection, more importantly, leave the dungeon in time.
The first line of each test case contains four integers W (1 <= W <= 50), H (1 <= H <= 50), L (1 <= L <= 1,000,000) and M (1 <= M <= 10). The dungeon is a rectangle area W block wide and H block high. L is the time limit, by which you need to reach the exit. You can move to one of the adjacent blocks up, down, left and right in each time unit, as long as the target block is inside the dungeon and is not a wall. Time starts at 1 when the game begins. M is the number of jewels in the dungeon. Jewels will be collected once the adventurer is in that block. This does not cost extra time.
The next line contains M integers,which are the values of the jewels.
The next H lines will contain W characters each. They represent the dungeon map in the following notation:
> [*] marks a wall, into which you can not move;
> [.] marks an empty space, into which you can move;
> [@] marks the initial position of the adventurer;
> [<] marks the exit stairs;
> [A] - [J] marks the jewels.
If the adventurer can make it to the exit stairs in the time limit, print the sentence "The best score is S.", where S is the maximum value of the jewels he can collect along the way; otherwise print the word "Impossible" on a single line.
4 4 2 2
100 200
****
*@A*
*B<*
****
4 4 1 2
100 200
****
*@A*
*B<*
****
12 5 13 2
100 200
************
*B.........*
*.********.*
*@...A....<*
************
The best score is 200.
Case 2:
Impossible
Case 3:
The best score is 300.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <queue>
#include <stdlib.h>
using namespace std; struct node
{
int x,y,cnt;
}que[]; int n,m,t,tj;
int cost[];
int dis[][];
char g[][];
int sid,tid;
int mlink[][];
int qf,qd;
int up[]={,,-,};
int rl[]={,,,-};
int dp[][][]; void bfs(int sx,int sy)
{
int mark[][];
memset(mark,,sizeof(mark));
qf=qd=;
node finode;
mark[sx][sy]=;
finode.x=sx; finode.y=sy; finode.cnt=;
que[qf++]=finode;
while(qf>qd)
{
node cur=que[qd++];
for(int i=;i<;i++)
{
int tx,ty;
tx=cur.x+up[i];
ty=cur.y+rl[i];
if( (tx>=&&tx<n)&&(ty>=&&ty<m) && g[tx][ty]!='*'&& mark[tx][ty]==)
{
mark[tx][ty]=;
node nwnode;
nwnode.cnt=cur.cnt+;
nwnode.x=tx;nwnode.y=ty;
que[qf++]=nwnode;
if(mlink[tx][ty]!=-)
{
dis[ mlink[tx][ty] ][ mlink[sx][sy] ]=nwnode.cnt;
dis[ mlink[sx][sy] ][ mlink[tx][ty] ]=nwnode.cnt;
}
}
}
}
} int main()
{
int T;
int tt=;
scanf("%d",&T);
int flag=;
while(T--)
{
if(flag) printf("\n");
flag=;
scanf("%d%d%d%d",&m,&n,&t,&tj);
for(int i=;i<tj;i++)
scanf("%d",cost+i);
for(int i=;i<n;i++)
{
scanf("%s",g[i]);
}
memset(dis,-,sizeof(dis));
memset(mlink,-,sizeof(mlink));
int id=;
for(int i=;i<n;i++)
for(int j=;j<m;j++)
{
if(g[i][j]!='.'&&g[i][j]!='*')
{
if(g[i][j]=='@')
mlink[i][j]=;
else if( g[i][j]=='<')
{
id++;
mlink[i][j]=tj+;
}
else mlink[i][j]=g[i][j]-'A'+;
}
}
if(id>=)
{
for(int i=;i<;i++)
printf("%d\n",cost[i]); }
for(int i=;i<n;i++)
for(int j=;j<m;j++)
if(g[i][j]!='.'&&g[i][j]!='<'&&g[i][j]!='*')
{
bfs(i,j);
}
////////////
memset(dp,-,sizeof(dp));
dp[][][]=; for(int i=;i<=tj;i++)
{
for(int j=;j<=tj;j++)
{
for(int k=; k<(<<(tj+)) ;k++)
{
if(dp[i-][j][k]!=- && dp[i-][j][k] < t )
{
for(int p=;p<=tj;p++)
{
if( (k&(<<p)) == &&dis[j][p]!=-)
{
if(dp[i][p][(k|(<<p))]==-) dp[i][p][k|(<<p)] = dp[i-][j][k]+dis[j][p];
else dp[i][p][(k|(<<p))] = min(dp[i][p][(k|(<<p))],dp[i-][j][k]+dis[j][p]);
}
}
}
}
}
}
int ans=-;
for(int i=;i<=tj;i++)
for(int j=;j<=tj;j++)
for(int k=;k<(<<(tj+));k++)
if(dp[i][j][k]!=-&&dis[j][tj+]!=-)
if(dp[i][j][k]+dis[j][tj+]<=t)
{
int tmp=;
for(int p=;p<=tj;p++)
if( ((<<p)&k)!= )
tmp+=cost[p-];
ans=max(ans,tmp);
} printf("Case %d:\n",tt++);
if(ans==-) printf("Impossible\n");
else printf("The best score is %d.\n",ans);
}
return ;
}
hdu 1044(bfs+状压)的更多相关文章
- hdu 2209 bfs+状压
http://acm.hdu.edu.cn/showproblem.php?pid=2209 不知为啥有种直觉.会出状压+搜索的题,刷几道先 简单的BFS.状压表示牌的状态, //#pragma co ...
- hdu 5025 bfs+状压
http://acm.hdu.edu.cn/showproblem.php?pid=5025 N*N矩阵 M个钥匙 K起点,T终点,S点需多花费1点且只需要一次,1-9表示9把钥匙,只有当前有I号钥匙 ...
- hdu 1429 bfs+状压
题意:这次魔王汲取了上次的教训,把Ignatius关在一个n*m的地牢里,并在地牢的某些地方安装了带锁的门,钥匙藏在地牢另外的某些地方.刚开始 Ignatius被关在(sx,sy)的位置,离开地牢的门 ...
- HDU 4771 BFS + 状压
Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 ...
- C - 小明系列故事――捉迷藏 HDU - 4528 bfs +状压 旅游-- 最短路+状压
C - 小明系列故事――捉迷藏 HDU - 4528 这个题目看了一下题解,感觉没有很难,应该是可以自己敲出来的,感觉自己好蠢... 这个是一个bfs 用bfs就很好写了,首先可以预处理出大明和二明能 ...
- hdu 5094 Maze (BFS+状压)
题意: n*m的迷宫.多多要从(1,1)到达(n,m).每移动一步消耗1秒.有P种钥匙. 有K个门或墙.给出K个信息:x1,y1,x2,y2,gi 含义是(x1,y1)与(x2,y2)之间有gi ...
- hdu 4771 Stealing Harry Potter's Precious (BFS+状压)
题意: n*m的迷宫,有一些格能走("."),有一些格不能走("#").起始点为"@". 有K个物体.(K<=4),每个物体都是放在& ...
- 孤岛营救问题 (BFS+状压)
https://loj.ac/problem/6121 BFS + 状压 写过就好想,注意细节debug #include <bits/stdc++.h> #define read rea ...
- HDU 5025:Saving Tang Monk(BFS + 状压)
http://acm.hdu.edu.cn/showproblem.php?pid=5025 Saving Tang Monk Problem Description <Journey to ...
随机推荐
- .NET设计模式(18):迭代器模式(Iterator Pattern)(转)
概述 在面向对象的软件设计中,我们经常会遇到一类集合对象,这类集合对象的内部结构可能有着各种各样的实现,但是归结起来,无非有两点是需要我们去关心的:一是集合内部的数据存储结构,二是遍历集合内部的数据. ...
- asp.net的sessionState节点详解
web.config关于sessionState节点的配置方案,sessionState有四种模式:off,inProc,StateServer,SqlServer. 1.off模式 从字面上就可以看 ...
- 采用Asp.Net的Forms身份验证时,持久Cookie的过期时间会自动扩展
原文:http://www.cnblogs.com/sanshi/archive/2012/06/22/2558476.html 若是持久Cookie,Cookie的有效期Expiration属性有当 ...
- JS面向(基于)对象编程--三大特征
抽象 在讲解面向对象编程的三大特征前,我们先了解什么叫抽象,在定义一个类时候,实际上就是把一类事物的共有的属性和行为提取出来,形成一个物理模型(模板).这种研究问题的方法称为抽象. 封装 什么是封装? ...
- [RM HA 1] Cloudera CDH5 RM HA功能验证
简介: 最新的Cloudera CDH5.0.0 beta版本已经支持RM的HA, 笔者为此简单验证了RM HA的功能. 后续将继续分析其HA的原理,以及其与社区RM HA的区别. 集群部属与RM f ...
- HTML基本操作
插入图片: 1.利用链接(静态) <img src="http://www.kmwzjs.com/useruploads/images/20101020_057600100825157 ...
- C++函数默认参数
C++中允许为函数提供默认参数,又名缺省参数. 使用默认参数时的注意事项: ① 有函数声明(原型)时,默认参数可以放在函数声明或者定义中,但只能放在二者之一 double sqrt(double f ...
- 网络打洞(P2P软件穿透内网进行通信) 原理
http://www.cnblogs.com/gansc23/archive/2010/10/20/1857066.html 首先先介绍一些基本概念:NAT(Network Address Trans ...
- JavaWeb项目开发案例精粹-第6章报价管理系统-002辅助类及配置文件
1. <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www ...
- Centos环境下部署游戏服务器-权限
部署Web服务器的时候,在"DocumentRoot"指向的根目录新建一个文件夹,然后将网页和资源放在这个文件夹里,通过地址http://192.168.0.100/Res/ind ...