Description face questions
All of us love treasures, right? That's why young Vasya is heading for a Treasure Island.
Treasure Island may be represented as a rectangular table n×m which is surrounded by the ocean. Let us number rows of the field with consecutive integers from 1 to n from top to bottom and columns with consecutive integers from 1 to m from left to right. Denote the cell in r-th row and c-th column as (r,c). Some of the island cells contain impassable forests, and some cells are free and passable. Treasure is hidden in cell (n,m).
Vasya got off the ship in cell (1,1). Now he wants to reach the treasure. He is hurrying up, so he can move only from cell to the cell in next row (downwards) or next column (rightwards), i.e. from cell (x,y) he can move only to cells (x+1,y) and (x,y+1). Of course Vasya can't move through cells with impassable forests.
Evil Witch is aware of Vasya's journey and she is going to prevent him from reaching the treasure. Before Vasya's first move she is able to grow using her evil magic impassable forests in previously free cells. Witch is able to grow a forest in any number of any free cells except cells (1,1) where Vasya got off his ship and (n,m) where the treasure is hidden.
Help Evil Witch by finding out the minimum number of cells she has to turn into impassable forests so that Vasya is no longer able to reach the treasure.
Input Format
First line of input contains two positive integers n, m (3≤n⋅m≤1000000), sizes of the island.
Following n lines contains strings si of length m describing the island, j-th character of string si equals "#" if cell (i,j) contains an impassable forest and "." if the cell is free and passable. Let us remind you that Vasya gets of his ship at the cell (1,1), i.e. the first cell of the first row, and he wants to reach cell (n,m), i.e. the last cell of the last row.
It's guaranteed, that cells (1,1) and (n,m) are empty.
Output Format
Print the only integer k, which is the minimum number of cells Evil Witch has to turn into impassable forest in order to prevent Vasya from reaching the treasure.
Sample data
Sample input
2 2
..
..
Sample Output
2
answer
Simple question ......
In fact, considering the maximum answer is the (1,2) and (2,1) plugged together, or (n-1, m) and (n, m-1) with stuffed, so the answer to a maximum of 2.
Then consider moves. If you simply determine that a neighbor is able to reach the start / finish line may appear the following conditions:
input:
. . . . . # .
. . . . . # .
. . . . . # .
. . . . . . .
# # # # # # .
. . . . . . .Then we find two adjacent nodes can reach the start and end, and the answer is 1, so we need other solutions.
Consider whether there are two paths. We first conduct a dfs, if there is no access directly output to zero. If this path will be blocked passageway, and a second dfs. If the answer to the second dfs has 2 output, otherwise the output 1.
About correctness, because of the nature dfs is not hit obstacles will not change direction (left-hand rule / right-hand rule) to move, even if there are two solutions, we will also give priority to plug does not have an impact on another answer the path can guarantee correct.
#include<bits/stdc++.h>
#define int long long
#define maxn 1000
using namespace std;
inline char get(){
static char buf[30000],*p1=buf,*p2=buf;
return p1==p2 && (p2=(p1=buf)+fread(buf,1,30000,stdin),p1==p2)?EOF:*p1++;
}
inline int read(){
register char c=getchar();register int f=1,_=0;
while(c>'9' || c<'0')f=(c=='-')?-1:1,c=getchar();
while(c<='9' && c>='0')_=(_<<3)+(_<<1)+(c^48),c=getchar();
return _*f;
}
int n,m;
set<pair<int,int> > mp,vis;
int dfs(int x,int y){
//cout<<x<<" "<<y<<endl;
if(x==n && y==m)return 1;
if(vis.count(make_pair(x,y)))return 0;
if(mp.count(make_pair(x,y)))return 0;
if(x>n || y>m)return 0;
vis.insert(make_pair(x,y));
if(dfs(x+1,y)==1){
mp.insert(make_pair(x+1,y));
return 1;
}
if(dfs(x,y+1)==1){
mp.insert(make_pair(x,y+1));
return 1;
}
return 0;
}
signed main(){
//freopen("1.txt","r",stdin);
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n>>m;
for(register int i=1;i<=n;i++){
char cas;
for(register int j=1;j<=m;j++){
cin>>cas;
if(cas=='#')mp.insert(make_pair(i,j));
}
}
if(dfs(1,1)==0){
cout<<0;
return 0;
}
vis.clear();
if(dfs(1,1)==0){
cout<<1;
return 0;
}
else{
cout<<2;
return 0;
}
return 0;
}