HDU 5791 Two DP

Two

 

Problem Description

 

Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A' is a subsequence of A. B' is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.

 

Input

 

The input contains multiple test cases.

For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.

 

Output

 

For each test case, output the answer mod 1000000007.

 

Sample Input

 

3 2 1 2 3 2 1 3 2 1 2 3 1 2

 

Sample Output

 

2 3

 

题意：

　　给你两个数组

　　问你有多少对公共子序列

题解：

　　设定dp[i][j] 表示以i结尾 j结尾的子序列的 答案数

　　n^2的转移

　　 假设当前为 a[i] == b[j] , 那么它可以继承的 就是 所有 的 i,j组合 +1 

　　 a[i] != b[j] 则 当前dp[i][j] = 0 咯

　　第一中继承 只需要利用前缀 优化就行

　　最后统计答案的话 就是sum[n][m]咯

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 1e3+10, M = 2e2+11, inf = 2e9, mod = 1e9+7;
typedef long long ll;

ll dp[N][N],sum[N][N],ans = 0;
int a[N],b[N],n,m;
int main()
{
    while (scanf("%d%d", &n, &m)!=EOF) {
        for (int i=1;i<=n;i++) scanf("%d", &a[i]);
        for (int i=1;i<=m;i++) scanf("%d", &b[i]);
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
            if (a[i]==b[j]) {
                    
                dp[i][j]=(sum[i-1][j-1]+1)%mod;
                
                sum[i][j]=(-sum[i-1][j-1]+dp[i][j]%mod+sum[i-1][j]%mod+sum[i][j-1])%mod;
            
            } else sum[i][j]=(-sum[i-1][j-1]%mod+sum[i-1][j]%mod+sum[i][j-1])%mod;
        ans=0;
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++) ans=(ans+dp[i][j])%mod;
        printf("%I64d
", (ans+mod)%mod);
    }
    return 0;
}