C. Ice Cave

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/540/problem/C

Description

You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.

The level of the cave where you are is a rectangular square grid of n rows and m columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.

Let's number the rows with integers from 1 to n from top to bottom and the columns with integers from 1 to m from left to right. Let's denote a cell on the intersection of the r-th row and the c-th column as (r, c).

You are staying in the cell (r1, c1) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (r2, c2) since the exit to the next level is there. Can you do this?

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 500) — the number of rows and columns in the cave description.

Each of the next n lines describes the initial state of the level of the cave, each line consists of m characters "." (that is, intact ice) and "X" (cracked ice).

The next line contains two integers, r1 and c1 (1 ≤ r1 ≤ n, 1 ≤ c1 ≤ m) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (r1, c1), that is, the ice on the starting cell is initially cracked.

The next line contains two integers r2 and c2 (1 ≤ r2 ≤ n, 1 ≤ c2 ≤ m) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.

Output

If you can reach the destination, print 'YES', otherwise print 'NO'.

Sample Input

4 6
X...XX
...XX.
.X..X.
......
1 6
2 2

Sample Output

YES

HINT

In the first sample test one possible path is:

After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended.

题意

有一个矩形区域,一开始你在一个地方,你只能走.的位置,走了之后.就会变成X,然后你必须让终点变成x,然后再走上去

问你可不可行

题解:

BFS搞一搞就好了,裸题= =

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //§ß§é§à§é¨f§³
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** string s[];
int dx[]={,-,,};
int dy[]={,,,-};
struct node
{
int x,y;
int t;
};
int vis[][];
int main()
{
int n=read(),m=read();
for(int i=;i<n;i++)
cin>>s[i];
node st,ed;
cin>>st.x>>st.y;
st.x--,st.y--;
st.t=;
cin>>ed.x>>ed.y;
ed.x--,ed.y--;
queue<node> q;
q.push(st);
while(!q.empty())
{
node now=q.front();
q.pop();
for(int i=;i<;i++)
{
node next=now;
next.x+=dx[i];
next.y+=dy[i];
next.t++;
if(next.x<||next.x>=n||next.y<||next.y>=m)
continue;
if(next.x==ed.x&&next.y==ed.y&&s[next.x][next.y]=='X')
{
printf("YES\n");
return ;
}
if(s[next.x][next.y]=='X')
continue;
q.push(next);
s[next.x][next.y]='X';
}
}
printf("NO\n");
}

Codeforces Round #301 (Div. 2) C. Ice Cave BFS的更多相关文章

  1. DFS/BFS Codeforces Round #301 (Div. 2) C. Ice Cave

    题目传送门 /* 题意:告诉起点终点,踩一次, '.'变成'X',再踩一次,冰块破碎,问是否能使终点冰破碎 DFS:如题解所说,分三种情况:1. 如果两点重合,只要往外走一步再走回来就行了:2. 若两 ...

  2. 贪心 Codeforces Round #301 (Div. 2) B. School Marks

    题目传送门 /* 贪心:首先要注意,y是中位数的要求:先把其他的都设置为1,那么最多有(n-1)/2个比y小的,cnt记录比y小的个数 num1是输出的1的个数,numy是除此之外的数都为y,此时的n ...

  3. 贪心 Codeforces Round #301 (Div. 2) A. Combination Lock

    题目传送门 /* 贪心水题:累加到目标数字的距离,两头找取最小值 */ #include <cstdio> #include <iostream> #include <a ...

  4. CodeForces Round #301 Div.2

    今天唯一的成果就是把上次几个人一起开房打的那场cf补一下. A. Combination Lock 此等水题看一眼样例加上那个配图我就明白题意了,可是手抽没有注释掉freopen,WA了一发. #in ...

  5. Codeforces Round #599 (Div. 2) D. 0-1 MST(bfs+set)

    Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers ...

  6. Codeforces Round #301 (Div. 2)(A,【模拟】B,【贪心构造】C,【DFS】)

    A. Combination Lock time limit per test:2 seconds memory limit per test:256 megabytes input:standard ...

  7. Codeforces Round #301 (Div. 2)A B C D 水 模拟 bfs 概率dp

    A. Combination Lock time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  8. 【解题报告】Codeforces Round #301 (Div. 2) 之ABCD

    A. Combination Lock 拨密码..最少次数..密码最多有1000位. 用字符串存起来,然后每位大的减小的和小的+10减大的,再取较小值加起来就可以了... #include<st ...

  9. 「日常训练」Ice Cave(Codeforces Round 301 Div.2 C)

    题意与分析(CodeForces 540C) 这题坑惨了我....我和一道经典的bfs题混淆了,这题比那题简单. 那题大概是这样的,一个冰塔,第一次踩某块会碎,第二次踩碎的会掉落.然后求可行解. 但是 ...

随机推荐

  1. javashop每次重新部署都要从新安装的问题

    javashop每次重新部署都要从新安装的问题 发现一个问题就是用MyEclipse是部署不上的,用eclipse才行. 这个问题的关键在于javashop有好多文件都是动态生成的,好多配置文件也是在 ...

  2. Windows降权

    使用invoke-tokenmanipulation进行降权 枚举所有令牌 PS C:\Users\SMC> Get-ExecutionPolicy Restricted PS C:\Users ...

  3. java中的matches -> 完全匹配

    matches是完全匹配.跟matcher不一样, matcher像perl正则, 能匹配到符合的都会返回true, 而这个matches要完全一模一样才行. import java.util.reg ...

  4. 64_s1

    SAASound-3.2-17.fc26.i686.rpm 13-Feb-2017 22:13 27650 SAASound-3.2-17.fc26.x86_64.rpm 13-Feb-2017 23 ...

  5. 一个好的Java时间工具类DateTime

    此类的灵感来源于C# 虽然网上有什么date4j,但是jar太纠结了,先给出源码,可以继承到自己的util包中,作为一个资深程序员,我相信都有不少好的util工具类,我也希望经过此次分享,能带动技术大 ...

  6. charles抓包误点deny处理办法及日常抓包

    误点deny方法在最底下~~ (博文为转载) 我们在开发网站项目的时候,我们可以通过浏览器的debug模式来看request以及response的数据,那么如果我们开发移动端项目没有网页呢?如何抓取数 ...

  7. Vue.js——60分钟快速入门(转)

    var vm = new Vue({ el: '#app', data: { people: [{ name: 'Jack', age: 30, sex: 'Male' }, { name: 'Bil ...

  8. 画弧线DrawArc的研究-我自己 -- 直线交接圆角

    procedure TForm4.Button7Click(Sender: TObject); var pwith: Integer; //画笔的宽度 hx1, hy1: Integer; //横线第 ...

  9. 几个python one-liner

    生成斐波那契数列的前10个数,从1开始.若生成前n个,改为range(n-2).代码很简单: List = reduce(lambda x, y: x + [x[-1] + x[-2]], range ...

  10. Webpack, VSCode 和 Babel 组件模块导入别名

    很多时候我们使用别人的库,都是通过 npm install,再简单的引入,就可以使用了.     1 2 import React from 'react' import { connect } fr ...