C. Ice Cave

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

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

Description

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.

//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[20];
//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=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
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("");
}
//**************************************************************************************string s[1000];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
struct node
{int x,y;int t;
};
int vis[1000][1000];
int main()
{int n=read(),m=read();for(int i=0;i<n;i++)cin>>s[i];node st,ed;cin>>st.x>>st.y;st.x--,st.y--;st.t=0;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=0;i<4;i++){node next=now;next.x+=dx[i];next.y+=dy[i];next.t++;if(next.x<0||next.x>=n||next.y<0||next.y>=m)continue;if(next.x==ed.x&&next.y==ed.y&&s[next.x][next.y]=='X'){printf("YES\n");return 0;}if(s[next.x][next.y]=='X')continue;q.push(next);s[next.x][next.y]='X';}}printf("NO\n");
}