hdu 5791 Two 二维dp

时间：2022-05-20 07:15:43

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 889    Accepted Submission(s): 405

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
Author
ZSTU
Source
 题意:给你a,b两个数组，可以从a数组中按顺序选出一些数字，然后b数组按顺序选出一些数字，要求
最后选出的数字按顺序相同，有多少选法。
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <vector>

#include <queue>

#include <cstring>

#include <string>

#include <algorithm>

using namespace std;

typedef  long long  ll;

typedef unsigned long long ull;

#define MM(a,b) memset(a,b,sizeof(a));

#define inf 0x7f7f7f7f

#define FOR(i,n) for(int i=1;i<=n;i++)

#define CT continue;

#define PF printf

#define SC scanf

const int mod=1000000007;

const int N=1000+10;

ull seed=13331;

ll dp[N][N];

int a[N],b[N];

int main()

{

    int n,m;

    while(~scanf("%d%d",&n,&m))

    {

        for(int i=1;i<=n;i++) scanf("%d",&a[i]);

        for(int i=1;i<=m;i++) scanf("%d",&b[i]);

        MM(dp,0);

        for(int i=1;i<=n;i++)

           for(int j=1;j<=m;j++)

                   {

                       dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;

                       if(a[i]==b[j]) dp[i][j]=(dp[i][j]+dp[i-1][j-1]+1)%mod;

                   }

        printf("%lld\n",dp[n][m]);

    }

    return 0;

}

　　分析：昨天比赛的时候一直想着怎么把暴力的复杂度降下去，想到了哈希，大数。。。就是将各种情况的对应到一个数值，，，，然并卵，，因为这道题目取的数字可以是不连续的，哈希跟大数就跪了。

正确解法：设dp[i][j]为当前枚举到a是i位,b是j位的时候，那么先撇开（i,j）不看，那么

dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]；（注意dp[i-1][j-1]多算了一次）

如果a[i]==b[j]，那么（i,j）还可以单独的产生组合，；

最后要注意%mod时，两者相减可能会产生负数

感觉取得数字要是不是连续的话一般就要上dp了

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