NOIP 模拟 $90\; \rm 回文$

题解 \(by\;zj\varphi\)

坐标 \(dp\)，朴素的想法是枚举从 \((1,1)\) 往右下走到的点，从 \((n,m)\) 往左上走到的点，复杂度 \(\mathcal O(\rm n^4)\)

优化一下，发现只有从左上向右下走的步数等于从右下向左上走的步数时才可能有贡献，所以枚举步数，同时枚举横向或纵向走的步数，另一方向的步数可以由这两个推出来，所以不用再枚举。

复杂度 \(\mathcal{O\rm(n)}\)

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        } 
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ll=long long;
    static const int N=507,MOD=993244853;
    int dp[2][N][N],n,m,al,hf,nw=1;
    char s[N][N];
    ll ans;
    inline int main() {
        FI=freopen("palin.in","r",stdin);
        FO=freopen("palin.out","w",stdout);
        scanf("%d%d",&n,&m);
        for (ri i(1);i<=n;pd(i)) scanf("%s",s[i]+1);
        dp[1][0][0]=s[1][1]==s[n][m];
        if (!dp[1][0][0]) return printf("0\n"),0;
        hf=(al=n+m-1)>>1;
        for (ri i(1);i<hf;pd(i)) {
            int cur=nw^1;
            memset(dp[cur],0,sizeof(dp[cur]));
            for (ri j(0);j<i;pd(j))
                for (ri k(0);k<i;pd(k)) {
                    if (j+1>m||i-j>n||n-i+k+1<1||m-k<1) continue;
                    if (j+2<=m&&m-k-1>0&&s[i-j][j+2]==s[n-i+k+1][m-k-1]) {
                        int &tmp=dp[cur][j+1][k+1];
                        tmp+=dp[nw][j][k];
                        if (tmp>=MOD) tmp-=MOD;
                    }
                    if (i-j+1<=n&&n-i+k>0&&s[i-j+1][j+1]==s[n-i+k][m-k]) {
                        int &tmp=dp[cur][j][k];
                        tmp+=dp[nw][j][k];
                        if (tmp>=MOD) tmp-=MOD;
                    }
                    if (j+2<=m&&n-i+k>0&&s[i-j][j+2]==s[n-i+k][m-k]) {
                        int &tmp=dp[cur][j+1][k];
                        tmp+=dp[nw][j][k];
                        if (tmp>=MOD) tmp-=MOD;
                    }
                    if (i-j+1<=n&&m-k-1>0&&s[i-j+1][j+1]==s[n-i+k+1][m-k-1]) {
                        int &tmp=dp[cur][j][k+1];
                        tmp+=dp[nw][j][k];
                        if (tmp>=MOD) tmp-=MOD;
                    }
                }
            nw=cur;
        }
        if (al&1) {
            for (ri i(1);i<=n;pd(i))
                for (ri j(1);j<=m;pd(j))
                    if (i-1+j-1==hf&&n-i+m-j==hf) {
                        if (i-1>0&&i+1<=n) ans+=dp[nw][j-1][m-j];
                        if (i-1>0&&j+1<=m) ans+=dp[nw][j-1][m-j-1];
                        if (j-1>0&&i+1<=n) ans+=dp[nw][j-2][m-j];
                        if (j-1>0&&j+1<=m) ans+=dp[nw][j-2][m-j-1];
                    }
        } else {
            for (ri i(0);i<hf&&i<=m-2;pd(i)) ans+=dp[nw][i][m-2-i];
            for (ri i(0);i<hf&&i<=m-1;pd(i)) ans+=dp[nw][i][m-1-i];
        }
        printf("%lld\n",ans%MOD);
        return 0;
    }
}
int main() {return nanfeng::main();}