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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=505;
//Code begin...

int dp[2][N][N];
char s[N][N];

int main ()
{
    int n=Scan(), m=Scan(), num=(n+m-2), flag=0;
    FOR(i,1,n) scanf("%s",s[i]+1);
    dp[flag][0][0]=1;
    FOR(i,0,num/2-1) {
        flag^=1; mem(dp[flag],0);
        FOR(j,0,min(i,m-1)) FOR(k,0,min(i,m-1)) {
            if (s[i-j+1][j+1]!=s[n-i+k][m-k]) continue;
            int tmp=dp[flag^1][j][k];
            if (j+1<m && m-k>1 && s[i-j+1][j+2]==s[n-i+k][m-k-1])
                dp[flag][j+1][k+1]=(dp[flag][j+1][k+1]+tmp)%MOD;
            if (j+1<m && n-i+k>1 && s[i-j+1][j+2]==s[n-i+k-1][m-k])
                dp[flag][j+1][k]=(dp[flag][j+1][k]+tmp)%MOD;
            if (i-j+1<n && m-k>1 && s[i-j+2][j+1]==s[n-i+k][m-k-1])
                dp[flag][j][k+1]=(dp[flag][j][k+1]+tmp)%MOD;
            if (i-j+1<n && n-i+k>1 && s[i-j+2][j+1]==s[n-i+k-1][m-k])
                dp[flag][j][k]=(dp[flag][j][k]+tmp)%MOD;
        }
    }
    int ans=0;
    if (num&1) FOR(j,0,min(num/2,m-1)) ans=(ans+dp[flag][j][m-2-j])%MOD, ans=(ans+dp[flag][j][num/2*2-n-j+2]);
    else FOR(j,0,min(num/2,m-1)) ans=(ans+dp[flag][j][m-1-j])%MOD;
    printf("%d
",ans);
    return 0;
}