Walk Out

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:
65536/65536 K (Java/Others)

                                                                                               Total Submission(s): 194    Accepted Submission(s): 32

Problem Description
In an n∗m maze,
the right-bottom corner is the exit (position (n,m) is
the exit). In every position of this maze, there is either a 0 or
a 1 written
on it.



An explorer gets lost in this grid. His position now is (1,1),
and he wants to go to the exit. Since to arrive at the exit is easy for him, he wants to do something more difficult. At first, he'll write down the number on position (1,1).
Every time, he could make a move to one adjacent position (two positions are adjacent if and only if they share an edge). While walking, he will write down the number on the position he's on to the end of his number. When finished, he will get a binary number.
Please determine the minimum value of this number in binary system.
 
Input
The first line of the input is a single integer T (T=10),
indicating the number of testcases. 



For each testcase, the first line contains two integers n and m (1≤n,m≤1000).
The i-th
line of the next n lines
contains one 01 string of length m,
which represents i-th
row of the maze.
 
Output
For each testcase, print the answer in binary system. Please eliminate all the preceding 0 unless
the answer itself is 0 (in
this case, print 0 instead).
 
Sample Input
2
2 2
11
11
3 3
001
111
101
 
Sample Output
111
101
 


题目大意:
       从(1,1)到(n,m),路径形成的二进制数最大。

解题思路:
      BFS+贪心,先是bfs找到离目标近期的距离。后用贪心让最前面的尽可能为0。策略就是每往前走一步,推断是否
能够是0,方法就是找它前面离它近期的能够取0的那一位,推断能否够从那个位置走到当前的位置。

代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=1000+100;
char s[maxn][maxn];
bool vis[maxn][maxn];
int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
struct node
{
int x;
int y;
};
int ans;
int n,m;
queue<node> q;
vector<node> son[maxn*2];
void BFS1()
{
memset(vis,false,sizeof(vis));
node p;
p.x=1;
p.y=1;
ans=0;
vis[1][1]=true;
while(!q.empty())
q.pop();
if(s[1][1]=='0')
{
ans=2;
q.push(p);
}
while(!q.empty())
{
node p2;
p=q.front();
q.pop();
if(p.x==n&&p.y==m)
{
if(s[n][m]=='0')
{
ans=n+m;
vis[n][m]=true;
}
break;
}
for(int i=0; i<4; i++)
{
p2.x=p.x+dir[i][0];
p2.y=p.y+dir[i][1];
if(p2.x>0&&p2.x<=n&&p2.y>0&&p2.y<=m&&!vis[p2.x][p2.y]&&s[p2.x][p2.y]=='0')
{
vis[p2.x][p2.y]=true;
if(p2.x+p2.y>ans)
{
ans=p2.x+p2.y;
}
q.push(p2);
}
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)
{
scanf("%s",s[i]+1);
}
BFS1();
if(ans==n+m)
printf("0\n");
else
{
for(int i=0; i<=n+m; i++)
son[i].clear();
int cur=0;//记录前一位0出现的位置
if(ans==0)//起点为1
ans=1;
else
{
for(int i=1; i<=n; i++)//找到全部的离(n,m)近期的点。
{
int j=ans-i;
if(j>=1&&j<=m&&vis[i][j]&&s[i][j]=='0')
{
node v;
v.x=i;
v.y=j;
son[ans].push_back(v);
}
}
cur=ans;
}
for(int i=ans+1; i<=n+m; i++)//枚举每一步
{
if(cur==0)//前面不存在0
{
for(int j=1; j<=n; j++)
{
int k=i-j;
node v;
if(k>=1&&k<=m&&s[j][k]=='0')
{
v.x=j;
v.y=k;
son[i].push_back(v);
cur=i;
}
}
}
else
{
for(int j=1; j<=n; j++)
{
int k=i-j;
node v;
if(k>=1&&k<=m&&s[j][k]=='0')
{
for(int l=0; l<son[cur].size(); l++)
{
v=son[cur][l];
if(v.x<=j&&v.y<=k&&i-cur>=j+k-v.x-v.y)//推断前面的0是否可达
{
node w;
w.x=j;
w.y=k;
son[i].push_back(w);
break;
}
}
}
}
if(son[i].size()>0)
cur=i;
} }
for(int i=ans+1; i<=n+m; i++)
{
if(son[i].size()>0)
printf("0");
else
printf("1");
}
printf("\n");
}
}
return 0;
}


     

hdu 5335 Walk Out (2015 Multi-University Training Contest 4)的更多相关文章

  1. hdu 5335 Walk Out (搜索)

    题目链接: hdu 5335 Walk Out 题目描述: 有一个n*m由0 or 1组成的矩形,探险家要从(1,1)走到(n, m),可以向上下左右四个方向走,但是探险家就是不走寻常路,他想让他所走 ...

  2. 2015 Multi-University Training Contest 4 hdu 5335 Walk Out

    Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Su ...

  3. HDU 5335 Walk Out(多校)

    Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Su ...

  4. hdu 5335 Walk Out 搜索+贪心

    Walk Out Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total S ...

  5. hdu 5335 Walk Out(bfs+寻找路径)

    Problem Description In an n∗m maze, the right-bottom corner or a written on it. An explorer gets los ...

  6. HDU 5335 Walk Out BFS 比较坑

    H - H Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status ...

  7. HDU 6091 - Rikka with Match | 2017 Multi-University Training Contest 5

    思路来自 某FXXL 不过复杂度咋算的.. /* HDU 6091 - Rikka with Match [ 树形DP ] | 2017 Multi-University Training Conte ...

  8. HDU 6125 - Free from square | 2017 Multi-University Training Contest 7

    思路来自这里 - - /* HDU 6125 - Free from square [ 分组,状压,DP ] | 2017 Multi-University Training Contest 7 题意 ...

  9. HDU 6129 - Just do it | 2017 Multi-University Training Contest 7

    比赛时脑子一直想着按位卷积... 按题解的思路: /* HDU 6129 - Just do it [ 规律,组合数 ] | 2017 Multi-University Training Contes ...

随机推荐

  1. EOJ 3348 树的顺序存储结构

    前面介绍了树的链式存储结构,那么如何用顺序存储来存储一棵树呢?在顺序存储时,我们除了存储每个结点值外,还要存储树中结点与结点之间的逻辑关系(即双亲与孩子结点之间的关系).下面介绍树的双亲存储法. 编号 ...

  2. js 数组包含

    function(arr,element){ return new RegExp('(^|,)'+element.toString()+'(,|$)').test(arr.toString()); }

  3. testNG中方法的调用顺序

    今天在执行selnium的test case时,总是遇到空指针错误.但是以前也有run成功过,然后换了各种方法定位元素,都失败了,所以怀疑应该不是元素定位不到的问题,所以可能是method之间有依赖, ...

  4. Windows环境下使用强大的wget工具

    安装 下载[http://www.interlog.com/~tcharron/wgetwin.html] 解压到目录 比如我解压到D:\Tool\wget 添加wget环境变量,这样使用就更方便了, ...

  5. Android网络编程随想录(1)

    本系列文章对整个Android网络编程进行了总结,包括基本的TCP/IP协议,HTTP协议,HTTPS协议,HttpClient,UrlConnection,一些网络通信的库到棉花糖新加入的OKHTT ...

  6. python课程设计笔记(四)整数、浮点数与字符串 time库

    整数类型(范围无限制) 十进制1 -1 二进制0b1 -0b1 八进制0o1 -0o1 十六进制0x1 -0x1 浮点类型(范围有限制但可忽略) 运算存在不确定尾数 :0.1+0.2!=0.3 原因: ...

  7. 【SQL】SELECT 语句

    1.1 SELECT基本语法: Select * |{[distinct]colum|expression [alias],…} from table; 1.2 查询当前用户所有在用的表及视图: HR ...

  8. spring boot的项目名称问题

    spring boot直接执行main函数时是不需要加项目名称,但是可以在application配置文件中配置server.context-path=/*来配置项目名称: spring mvc的运行方 ...

  9. AngularJS指令进阶 -- ngModelController详解

    大家都知道AngularJS中的指令是其尤为复杂的一个部分,但是这也是其比较好玩的地方.这篇文章我们就来说一说如何在我们自定义的指令中,利用ngModel的controller来做双向数据绑定,本文对 ...

  10. MVC 接收文件

    [HttpPost] public ActionResult Layedit() { var files = Request.Files; //获得所上传的所有文件 ) { HttpPostedFil ...