NC17621 管道取珠(模型转换+DP)

题目链接

题意：







题解：

AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1024523;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int n,m;
int dp[2][510][510];
string a,b;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    cin>>n>>m;
    cin>>a>>b;a='.'+a,b='.'+b;
    dp[0][0][0]=1;
    int cur=0;
    for(int i=0;i<=n;i++,cur^=1)
        for(int j=0;j<=m;j++)
            for(int k=0;k<=n;k++){
                int l=i+j-k;
                int &x=dp[cur][j][k];
                if(l<0||l>m)continue;
                if(a[i+1]==a[k+1])dp[cur^1][j][k+1]=(dp[cur^1][j][k+1]+x)%mod;
                if(a[i+1]==b[l+1])dp[cur^1][j][k]=(dp[cur^1][j][k]+x)%mod;
                if(b[j+1]==a[k+1])dp[cur][j+1][k+1]=(dp[cur][j+1][k+1]+x)%mod;
                if(b[j+1]==b[l+1])dp[cur][j+1][k]=(dp[cur][j+1][k]+x)%mod;
                x=0;
            }
    cout<<dp[cur][m][n];
    return 0;
}