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 ;
}