Two HDU - 5791 (DP)

最新推荐文章于 2020-02-21 21:48:38 发布

原创最新推荐文章于 2020-02-21 21:48:38 发布 · 355 阅读

1 ·

CC 4.0 BY-SA版权

动态规划专栏收录该内容

40 篇文章

订阅专栏

本文探讨了一道经典的动态规划问题——求两个序列的最长公共子序列的数量，并提供了一个具体的实现示例。通过状态转移方程，有效地解决了该问题。

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)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
dp题，原型是最长公共子序列，只不过这个要求个数。
状态：

d p [x] [y] : a 序 列 1 到 x 和 b 序 列 1 到 y 符 合 要 求 的 答 案

$dp[x][y]:a序列1到x和b序列1到y符合要求的答案$
状态转移方程：

当 a [x] = b [y] ， 则 d p [x] [y] = d p [x] [y - 1] + d p [x - 1] [y] + 1

$当a[x]=b[y]，则dp[x][y]=dp[x][y-1]+dp[x-1][y]+1$

当 不 等 ， d p [x] [y] = d p [x] [y - 1] + d p [x - 1] [y] - d p [x - 1] [y - 1]

$当不等，dp[x][y]=dp[x][y-1]+dp[x-1][y]-dp[x-1][y-1]$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mod 1000000007
using namespace std;
int n,m;
int dp[1003][1003];
int a[1003],b[1003];
int solve(int x,int y)
{
    if(dp[x][y]!=-1)
        return dp[x][y];
    if(a[x]==b[y])
        return dp[x][y]=(solve(x,y-1)+solve(x-1,y)+1)%mod;
    int temp=solve(x,y-1)+solve(x-1,y)-solve(x-1,y-1);
    if(temp<0)
        return dp[x][y]=(temp+mod)%mod;
    else
        return dp[x][y]=temp%mod;
}
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=m;i++)
            scanf("%d",b+i);
        memset(dp,-1,sizeof(dp));
            for(int i=0;i<=1000;i++)
        dp[i][0]=0,dp[0][i]=0;
        printf("%d\n",solve(n,m));
    }
    return 0;
}