[Usaco 2015 open] palindromic paths

[Question link]

Https://www.lydsy.com/JudgeOnline/problem.php? Id = 4098

[Algorithm]

Obviously, the position (x, y) of the I-th letter in the return path must meet the following requirements: X + Y-1 = I

Use F [I] [J] [k] to represent the current step I. The upper left X axis is J, and the lower right X axis is K, how many solutions are there to make the letter sequence on both paths the same, DP is enough

Time Complexity: O (N ^ 3)

Scroll array to optimize the space complexity to O (N ^ 2)

[Code]

#include<bits/stdc++.h>using namespace std;#define MAXN 510 const int P = 1e9 + 7;int n;int f[2][MAXN][MAXN];char mp[MAXN][MAXN];inline void update(int &x , int y){        x += y;        x %= P;}inline bool valid(int x , int y){        return x >= 1 && x <= n && y >= 1 && y <= n;}int main(){                scanf("%d",&n);        for (int i = 1; i <= n; i++) scanf("%s",mp[i] + 1);        if (mp[1][1] == mp[n][n])        {                f[1][1][n] = 1;            } else         {                printf("0\n");                return 0;        }        for (int i = 1; i <= n; i++)        {                int now = i & 1 , nxt = now ^ 1;                memset(f[nxt] , 0 , sizeof(f[nxt]));                for (int x = 1; x <= n; x++)                {                        for (int y = 1; y <= n; y++)                        {                                int t1 = i + 1 - x , t2 = 2 * n - y + 1 - i;                                if (!valid(x , t1) || !valid(y , t2)) continue;                                if (!f[now][x][y]) continue;                                if (valid(x , t1 + 1))                                {                                        if (valid(y , t2 - 1) && mp[x][t1 + 1] == mp[y][t2 - 1])                                                 update(f[nxt][x][y] , f[now][x][y]);                                        if (valid(y - 1 , t2) && mp[x][t1 + 1] == mp[y - 1][t2])                                                update(f[nxt][x][y - 1] , f[now][x][y]);                                }                                    if (valid(x + 1 , t1))                                {                                        if (valid(y , t2 - 1) && mp[x + 1][t1] == mp[y][t2 - 1])                                                update(f[nxt][x + 1][y] , f[now][x][y]);                                        if (valid(y - 1 , t2) && mp[x + 1][t1] == mp[y - 1][t2])                                                update(f[nxt][x + 1][y - 1] , f[now][x][y]);                                 }                        }                    }                }        int ans = 0;        for (int i = 1; i <= n; i++) update(ans , f[n & 1][i][i]);        printf("%d\n" , ans);                return 0;    }

 

[Usaco 2015 open] palindromic paths