Problem

给你2*n的格子，每个格子有一个字母，从任意一点出发，不重复的经过上下左右，生成要求的字符串。问有几种不同的走法。

Solution

分三段，左U型、中间、右U型。
分别枚举左边和右边的长度，中间一段用Dp来解决。
Dp[i][j][k]，i,j,k表示当前在(i,j)位置，枚举到第k个字符。

Notice

特殊情况下有重复。

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, mod = 1e9 + 7, Ha = 826036489, N = 2000;
const double eps = 1e-6, phi = acos(-1.0);
ll modd(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
char S[2][N + 5], st[N + 5];
ll f[2][N + 5][N + 5], mi[N + 5];
int n, m;
void Calc(ll &X, ll Y)
{
    X += Y;
    if (X >= mod) X -= mod;
}
struct Hash
{
    ll hash[N + 5];
    void Make(int n, char *s)
    {
        hash[0] = 0;
        rep(i, 1, n) hash[i] = (hash[i - 1] * 31 + s[i] - 'a') % Ha;
    }
    ll Cut(int l, int r)
    {
        return (hash[r] - hash[l - 1] * mi[r - l + 1] % Ha + Ha) % Ha;
    }
}pre[2], suf[2], Comp;
ll Solve(int flag)
{
    ll T = 0;
    memset(f, 0, sizeof(f));
    rep(j, 1, n)
    {
        f[0][j][0] = f[1][j][0] = 1;
        rep(i, 0, 1)
        {
            rep(k, 2, min(n - j + 1, m / 2))
                if (Comp.Cut(m - 2 * k + 1, m - k) == pre[i].Cut(j, j + k - 1) && Comp.Cut(m - k + 1, m) == suf[1 - i].Cut(n - (j + k - 1) + 1, n - j + 1))
                    if (2 * k != m || flag) Calc(T, f[i][j][m - 2 * k]);
            rep(k, 2, min(j, m / 2))
                if (Comp.Cut(k + 1, 2 * k) == pre[i].Cut(j - k + 1, j) && Comp.Cut(1, k) == suf[1 - i].Cut(n - j + 1, n - (j - k + 1) + 1))
                    if (2 * k != m || flag) Calc(f[i][j + 1][2 * k], 1);
        }
        rep(i, 0, 1)
            rep(k, 0, m - 1)
                if (S[i][j] == st[k + 1])
                {
                    Calc(f[i][j + 1][k + 1], f[i][j][k]);
                    if (k + 2 <= m && S[1 - i][j] == st[k + 2])
                        Calc(f[1 - i][j + 1][k + 2], f[i][j][k]);
                }
        rep(i, 0, 1) Calc(T, f[i][j + 1][m]);
    }
    return T;
}
int sqz()
{
    scanf("%s%s%s", S[0] + 1, S[1] + 1, st + 1);
    n = strlen(S[0] + 1), m = strlen(st + 1);
    mi[0] = 1;
    rep(i, 1, 2000) mi[i] = (mi[i - 1] * 31) % Ha;
    rep(i, 0, 1)
    {
        pre[i].Make(n, S[i]);
        reverse(S[i] + 1, S[i] + n + 1);
        suf[i].Make(n, S[i]);
        reverse(S[i] + 1, S[i] + n + 1);
    }
    Comp.Make(m, st);
    if (m == 1)
    {
        printf("%I64d\n", Solve(1) % mod);
        return 0;
    }
    ll ans = 0;
    Calc(ans, Solve(1));
    reverse(st + 1, st + m + 1);
    Comp.Make(m, st);
    Calc(ans, Solve(0));
    if (m == 2)
        rep(i, 1, n)
        {
            if (S[0][i] == st[1] && S[1][i] == st[2]) Calc(ans, mod - 1);
            if (S[1][i] == st[1] && S[0][i] == st[2]) Calc(ans, mod - 1);
        }
    printf("%I64d\n", ans);
    return 0;
}