HDU 5791：Two（DP）

http://acm.hdu.edu.cn/showproblem.php?pid=5791

 

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：有两个串，求两两子串相同的个数有多少（可以不连续）。

思路：有点类似于LCS的DP，（+MOD）%MOD是因为有可能减出负数，因为取MOD一开始可能很大，后面变得很小。

 

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
#define N 1005
#define MOD 1000000007
typedef long long LL;

LL dp[N][N];
int a[N], b[N];
/*
1 2 3
2 1
*/
int main()
{
    int n, m;
    while(~scanf("%d%d", &n, &m)) {
        dp[0][0] = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", a+i);
            dp[i][0] = 0;
        }
        for(int i = 1; i <= m; i++) {
            scanf("%d", b+i);
            dp[0][i] = 0;
        }
/*
有这三部分
dp[i-1][j-1]
dp[i-1][j] - dp[i-1][j-1]
dp[i][j-1] - dp[i-1][j-1]
如果不匹配的话 dp[i][j] = dp[i-1][j] - dp[i-1][j-1] + dp[i][j-1] - dp[i-1][j-1] + dp[i-1][j-1]
匹配的话 dp[i][j] = 不匹配的状态 + dp[i-1][j-1] + 1
dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] 表示当前不匹配的状态有多少种，因为dp[i-1][j]和dp[i][j]中有dp[i-1][j-1]重复，所以要减去一个
如果当前匹配的话，就不用减去，因为要留一个来和当前的a[i]和b[j]匹配。
*/
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                if(a[i] == b[j]) {
                    dp[i][j] = (dp[i-1][j] + dp[i][j-1] + 1 + MOD) % MOD;
                } else {
                    dp[i][j] = (dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1] + MOD) % MOD;
                }
            }
        }

        printf("%I64d\n", dp[n][m] % MOD);
    }
    return 0;
}