BZOJ 1566 管道取珠(DP)

求方案数的平方之和。这个看起来很难解决。如果转化为求方案数的有序对的个数。那么就相当于求A和B同时取，最后序列一样的种数。

令dp[i][j][k]表示A在上管道取了i个，下管道取了j个，B在上管道取了k个，下管道取了i+j-k个珠子的序列相同的种数。

那么状态转移方程显然可得。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

//# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

int dp[N][N][N];

char s1[N], s2[N];

int dfs(int x, int y, int z){

    if (~dp[x][y][z]) return dp[x][y][z];

    int res=;

    if (x>&&z>&&s1[x]==s1[z]) res+=dfs(x-,y,z-);

    if (x>&&x+y>z&&s1[x]==s2[x+y-z]) res+=dfs(x-,y,z);

    if (y>&&z>&&s2[y]==s1[z]) res+=dfs(x,y-,z-);

    if (y>&&x+y>z&&s2[y]==s2[x+y-z]) res+=dfs(x,y-,z);

    return dp[x][y][z]=res%MOD;

}

int main ()

{

    int n, m;

    scanf("%d%d%s%s",&n,&m,s1+,s2+);

    mem(dp,-); dp[][][]=;

    printf("%d\n",dfs(n,m,n));

    return ;

}