51nod 1503 猪和回文(多线程DP)

虚拟两个点，一个从左上角开始走，一个从右下角开始走，定义dp[i][j][k]表示走了i步后，第一个点横向走了j步，第二个点横向走了k步后形成的回文方法种数。

转移方程显然可得，然后滚动数组搞一搞。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# 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 FDR(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;

inline 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;

}

inline void Out(int a) {

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

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

    putchar(a%+'');

}

const int N=;

//Code begin...

int dp[][N][N];

char s[N][N];

int main ()

{

    int n=Scan(), m=Scan(), num=(n+m-), flag=;

    FOR(i,,n) scanf("%s",s[i]+);

    dp[flag][][]=;

    FOR(i,,num/-) {

        flag^=; mem(dp[flag],);

        FOR(j,,min(i,m-)) FOR(k,,min(i,m-)) {

            if (s[i-j+][j+]!=s[n-i+k][m-k]) continue;

            int tmp=dp[flag^][j][k];

            if (j+<m && m-k> && s[i-j+][j+]==s[n-i+k][m-k-])

                dp[flag][j+][k+]=(dp[flag][j+][k+]+tmp)%MOD;

            if (j+<m && n-i+k> && s[i-j+][j+]==s[n-i+k-][m-k])

                dp[flag][j+][k]=(dp[flag][j+][k]+tmp)%MOD;

            if (i-j+<n && m-k> && s[i-j+][j+]==s[n-i+k][m-k-])

                dp[flag][j][k+]=(dp[flag][j][k+]+tmp)%MOD;

            if (i-j+<n && n-i+k> && s[i-j+][j+]==s[n-i+k-][m-k])

                dp[flag][j][k]=(dp[flag][j][k]+tmp)%MOD;

        }

    }

    int ans=;

    if (num&) FOR(j,,min(num/,m-)) ans=(ans+dp[flag][j][m--j])%MOD, ans=(ans+dp[flag][j][num/*-n-j+]);

    else FOR(j,,min(num/,m-)) ans=(ans+dp[flag][j][m--j])%MOD;

    printf("%d\n",ans);

    return ;

}