hdu5800 To My Girlfriend dp 需要比较扎实的dp基础。

To My Girlfriend

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1288    Accepted Submission(s): 492

Problem Description

Dear Guo

I never forget the moment I met with you.You carefully asked me: "I have a very difficult problem. Can you teach me?".I replied with a smile, "of course"."I have n items, their weight was a[i]",you said,"Let's define f(i,j,k,l,m) to be the number of the subset of the weight of n items was m in total and has No.i and No.j items without No.k and No.l items.""And then," I asked.You said:"I want to know

∑i=1n∑j=1n∑k=1n∑l=1n∑m=1sf(i,j,k,l,m)(i,j,k,laredifferent)

Sincerely yours,
Liao

 

 

Input

The first line of input contains an integer T(T≤15) indicating the number of test cases.
Each case contains 2 integers n, s (4≤n≤1000,1≤s≤1000). The next line contains n numbers: a1,a2,…,an (1≤ai≤1000).

 

 

Output

Each case print the only number — the number of her would modulo 109+7 (both Liao and Guo like the number).

 

 

Sample Input

2 4 4 1 2 3 4 4 4 1 2 3 4

 

 

Sample Output

8 8

 

 

Author

UESTC

 

 

Source

2016 Multi-University Training Contest 6

/**
题目：To My Girlfriend
链接：http://acm.hdu.edu.cn/showproblem.php?pid=5800
题意：如原题公式所示。
思路：

来源出题方给的题解。

令dp[i][j][s1][s2]表示前i个物品填了j的体积，有s1个物品选为必选，s2个物品选为必不选的方案数
(0<=s1,s2<=2)，则有转移方程
dp[i][j][s1][s2] = dp[i - 1][j][s1][s2] + dp[i-1][j-a[i]][s1][s2] + dp[i - 1][j - a[i]][s1 - 1][s2] + dp[i - 1][j][s1][s2 - 1]，
边界条件为dp[0][0][0][0] = 1，时间复杂度O(NS*3^2)。


dp[i - 1][j][s1][s2]： 不选第i个
dp[i-1][j-a[i]][s1][s2]： 选第i个
dp[i - 1][j - a[i]][s1 - 1][s2]： 第i个必选
dp[i - 1][j][s1][s2 - 1]： 第i个必不选


最终结果为ans += dp[n][x][2][2]*4;(1<=x<=s)
因为：
dp[n][x][2][2]算出来的都是没有排列时候选的i,j,k,l;
经过排列即：(i,j),(j,i),(k,l),(l,k)共四种。所有*4；
*/

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e6+5;
const double eps = 1e-12;
int dp[1001][1001][3][3];
int a[1004];
int n, s;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&s);
        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);
        memset(dp, 0, sizeof dp);
        dp[0][0][0][0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 0; j <= s; j++){
                for(int s1 = 0; s1 <= 2; s1++){
                    for(int s2 = 0; s2 <= 2; s2++){
                        dp[i][j][s1][s2] = dp[i-1][j][s1][s2];
                        if(s1!=0&&j>=a[i]){
                            dp[i][j][s1][s2] += dp[i-1][j-a[i]][s1-1][s2];
                            dp[i][j][s1][s2] %= mod;
                        }
                        if(j>=a[i]){
                            dp[i][j][s1][s2] += dp[i-1][j-a[i]][s1][s2];
                            dp[i][j][s1][s2] %= mod;
                        }
                        if(s2!=0){
                            dp[i][j][s1][s2] += dp[i-1][j][s1][s2-1];
                            dp[i][j][s1][s2] %= mod;
                        }
                    }
                }
            }
        }
        LL ans = 0;
        for(int i = 1; i <= s; i++) {
            ans = (ans+dp[n][i][2][2])%mod;
        }
        printf("%lld\n",ans*4%mod);
    }
    return 0;
}