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-)*m+j;
}

int main(){
    cin>>n>>m;
    for(int i=;i<=n;i++)
        for(int j=;j<=m;j++)
            scanf("\n%c",&mp[id(i,j)]);

    stk.push(Node(,));
    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)]=;

        if(x+<=n && y<=m && mp[id(x+,y)]=='.')
            stk.push(Node(x+,y));
        if(x<=n && y+<=m && mp[id(x,y+)]=='.')
            stk.push(Node(x,y+));
    }

    if(!vis[id(n,m)]){puts("");return ;}

    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-> && y> && vis[id(x-,y)])
            stk.push(Node(x-,y));
        else if(x> && y-> && vis[id(x,y-)])
            stk.push(Node(x,y-));
    }

    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> && y-> && vis[id(x,y-)])
            stk.push(Node(x,y-));
        else if(x-> && y> && vis[id(x-,y)])
            stk.push(Node(x-,y));
    }
    for(int i=;i<=n;i++)
        for (int j=;j<=m;j++){
            if(i== && j==)continue;
            if(i==n && j==m)continue;
            if(path[id(i,j)]==){
                puts("");
                return ;
            }
        }
    puts("");

}