CF:1214D.Treasure Island(有向图必经点)

题目链接
可以挡住起点旁两个格子，故答案最多为$222$。若$(1,1)$到不了$(n,m)$，则答案为0。否则记$F1[x][y]$为$(1,1)$到$(x,y)$的路径数，$F2[x][y]$为$(x,y)$到$(n,m)$的路径数。如果有一个$(x,y)$满足$F1[x][y]\ast F2[x][y]=F1[n][m]$，则$(x,y)$是必经点，挡住它即可，答案为1；否则答案为2。写个DP取一个较大的模数即可。

该题直接用map会超时,所以要将二维转换成一维的形式

#include <bits/stdc++.h>
using namespace std;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef unsigned long long ULL;
typedef pair<double,double> PDD;
typedef long long LL;
typedef pair<LL,LL> PII;
#define int LL
//#define int ULL
//#define x first
#define y second
const int inf=0x3f3f3f3f;
//const int mod=1e9+9;
const int mod=998244353;
inline int read(){
    int s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-') w=-1;
        ch=getchar();
    }
    while(ch<='9'&&ch>='0'){
        s=(s<<1)+(s<<3)+(ch-'0');
        ch=getchar();
    }
    return s*w;
}
const int maxn=2e6+5;
int n,m;
char maps[maxn];
int f[maxn],g[maxn];
bool flag[maxn];
int get(int x,int y){
    return (x-1)*m+y;
}
bool dfs(int x,int y){
    if(x==0||x==n+1||y==0||y==m+1) return false;
    if(maps[get(x,y)]=='#') return false;
    if(flag[get(x,y)]) return false;
    if(x==n&&y==m) return true;
    flag[get(x,y)]=true;
    bool is_flag=false;
    is_flag|=dfs(x+1,y);
    is_flag|=dfs(x,y+1);
    return is_flag;
}
void solve(){
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            cin>>maps[get(i,j)];
    if(!dfs(1,1)){
        cout<<0<<'\n';
        return ;
    }
    f[1]=g[n*m]=1;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            if(maps[get(i,j)]=='#') continue;
            if(i==1&&j==1) continue;
            f[get(i,j)]=(f[get(i-1,j)]+f[get(i,j-1)])%mod;
        }
    for(int i=n;i>=1;i--)
        for(int j=m;j>=1;j--){
            if(i==n&&j==m) continue;
            if(maps[get(i,j)]=='#') continue;
            g[get(i,j)]=(g[get(i+1,j)]+g[get(i,j+1)])%mod;
        }
    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((f[get(i,j)]*g[get(i,j)])%mod==f[get(n,m)]){
                cout<<1<<'\n';
                return ;
            }
        }
    }
    cout<<2<<'\n';
}
signed main(){
    ios;
    int t;t=1;
    while(t--) solve();
    return 0;
}

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