网格图必经点+dfs——cf1214D

先正着走一次把所有可行路径标记出来，然后倒着走两条路径，一条是能向下就向下的路径，另一条能向右就向右。

如果这两条路径相交，那么(1,1)-(n,m)路径上比有个必经点，把这个必经点封上，答案是1，如果没有必经点答案是2，如果(1,1)-(n,m)不连通，答案是0

直接用递归dfs会爆栈。所以要用栈来代替递归

#include<bits/stdc++.h>
using namespace std;
#define N 1000005
char mp[N];
int n,m;
struct Node{
    int x,y;
    Node(){}
    Node(int x,int y):x(x),y(y){}
};
stack<Node>stk;
int vis[N],path[N];
int id(int i,int j){return (i-1)*m+j;
}
 
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("
%c",&mp[id(i,j)]);
    
    stk.push(Node(1,1));
    while(stk.size()){
        Node c=stk.top();stk.pop();
        int x=c.x,y=c.y;
        if(vis[id(x,y)])continue;
        vis[id(x,y)]=1;
        
        if(x+1<=n && y<=m && mp[id(x+1,y)]=='.')
            stk.push(Node(x+1,y));
        if(x<=n && y+1<=m && mp[id(x,y+1)]=='.')
            stk.push(Node(x,y+1));    
    }
    
    if(!vis[id(n,m)]){puts("0");return 0;}
    
    while(stk.size())stk.pop();
    stk.push(Node(n,m));
    while(stk.size()){
        Node c=stk.top();stk.pop();
        int x=c.x,y=c.y;
        path[id(x,y)]++;
        
        if(x-1>0 && y>0 && vis[id(x-1,y)])
            stk.push(Node(x-1,y));
        else if(x>0 && y-1>0 && vis[id(x,y-1)])
            stk.push(Node(x,y-1));
    }
    
    while(stk.size())stk.pop();
    stk.push(Node(n,m));
    while(stk.size()){
        Node c=stk.top();stk.pop();
        int x=c.x,y=c.y;
        path[id(x,y)]++;
        
        if(x>0 && y-1>0 && vis[id(x,y-1)])
            stk.push(Node(x,y-1));
        else if(x-1>0 && y>0 && vis[id(x-1,y)])
            stk.push(Node(x-1,y));
    }
    for(int i=1;i<=n;i++)
        for (int j=1;j<=m;j++){
            if(i==1 && j==1)continue;
            if(i==n && j==m)continue;
            if(path[id(i,j)]==2){
                puts("1");
                return 0;
            }
        }
    puts("2");
    
}