HDU5800 To My Girlfrend（计数dp）

To My Girlfriend

传送门1
传送门2

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,l are different)

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

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题意

给定n个数，其中选定若干数，这若干数的权值和为m，且这些数中没有下标为i,j的数，有下标为k,l的数的集合个数。

分析

定义dp[i][j][ii][jj]表示前i个物品，总权值为j，已有ii个必选，jj必不选的方案数。
显然i<=n,j<=s,0<=ii,jj<=2
当然对于当前一个状态，它有四种转移状态：
1. 　选中当前的，　增加权值，　增加必　选个数。
2. 　选中当前的，　增加权值，不增加必　选个数。
3. 不选中当前的，不增加权值，　增加必不选个数。
4. 不选中当前的，不增加权值，不增加必不选个数。
在f(i,j,k,l,m)中，因为i,j可以互换，l,k也可以互换，故而最后方案数乘以4即为所求。
参考

CODE

#include<cstdio>#include<memory.h>#define mod 1000000007#define N 1005#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)int dp[N][N][3][3],a[N];int main() {    int T,n,s,tmp;    int ans;    scanf("%d",&T);    while(T--) {        ans=0;        scanf("%d%d",&n,&s);        FOR(i,1,n)scanf("%d",&a[i]);        memset(dp,0,sizeof dp);        dp[1][a[1]][0][0]=1;        dp[1][a[1]][1][0]=1;        dp[1][0][0][0]=1;        dp[1][0][0][1]=1;        FOR(i,2,n)FOR(j,0,s) {            tmp=j+a[i];            if(tmp<=s) { //卡常大法好                dp[i][tmp][0][0]=(dp[i][tmp][0][0]+dp[i-1][j][0][0])%mod;                dp[i][tmp][1][0]=(dp[i][tmp][1][0]+dp[i-1][j][0][0])%mod;                dp[i][tmp][1][0]=(dp[i][tmp][1][0]+dp[i-1][j][1][0])%mod;                dp[i][tmp][2][0]=(dp[i][tmp][2][0]+dp[i-1][j][1][0])%mod;                dp[i][tmp][2][0]=(dp[i][tmp][2][0]+dp[i-1][j][2][0])%mod;                dp[i][tmp][0][1]=(dp[i][tmp][0][1]+dp[i-1][j][0][1])%mod;                dp[i][tmp][1][1]=(dp[i][tmp][1][1]+dp[i-1][j][0][1])%mod;                dp[i][tmp][1][1]=(dp[i][tmp][1][1]+dp[i-1][j][1][1])%mod;                dp[i][tmp][2][1]=(dp[i][tmp][2][1]+dp[i-1][j][1][1])%mod;                dp[i][tmp][2][1]=(dp[i][tmp][2][1]+dp[i-1][j][2][1])%mod;                dp[i][tmp][0][2]=(dp[i][tmp][0][2]+dp[i-1][j][0][2])%mod;                dp[i][tmp][1][2]=(dp[i][tmp][1][2]+dp[i-1][j][0][2])%mod;                dp[i][tmp][1][2]=(dp[i][tmp][1][2]+dp[i-1][j][1][2])%mod;                dp[i][tmp][2][2]=(dp[i][tmp][2][2]+dp[i-1][j][1][2])%mod;                dp[i][tmp][2][2]=(dp[i][tmp][2][2]+dp[i-1][j][2][2])%mod;            }            dp[i][j][0][0]=(dp[i][j][0][0]+dp[i-1][j][0][0])%mod;            dp[i][j][0][1]=(dp[i][j][0][1]+dp[i-1][j][0][0])%mod;            dp[i][j][0][1]=(dp[i][j][0][1]+dp[i-1][j][0][1])%mod;            dp[i][j][0][2]=(dp[i][j][0][2]+dp[i-1][j][0][1])%mod;            dp[i][j][0][2]=(dp[i][j][0][2]+dp[i-1][j][0][2])%mod;            dp[i][j][1][0]=(dp[i][j][1][0]+dp[i-1][j][1][0])%mod;            dp[i][j][1][1]=(dp[i][j][1][1]+dp[i-1][j][1][0])%mod;            dp[i][j][1][1]=(dp[i][j][1][1]+dp[i-1][j][1][1])%mod;            dp[i][j][1][2]=(dp[i][j][1][2]+dp[i-1][j][1][1])%mod;            dp[i][j][1][2]=(dp[i][j][1][2]+dp[i-1][j][1][2])%mod;            dp[i][j][2][0]=(dp[i][j][2][0]+dp[i-1][j][2][0])%mod;            dp[i][j][2][1]=(dp[i][j][2][1]+dp[i-1][j][2][0])%mod;            dp[i][j][2][1]=(dp[i][j][2][1]+dp[i-1][j][2][1])%mod;            dp[i][j][2][2]=(dp[i][j][2][2]+dp[i-1][j][2][1])%mod;            dp[i][j][2][2]=(dp[i][j][2][2]+dp[i-1][j][2][2])%mod;        }        FOR(m,1,s)ans=(ans+dp[n][m][2][2])%mod;        printf("%d\n",(4LL*ans)%mod);    }    return 0;}

当然你不想看卡常的话可以看下面的代码

#include<cstdio>#include<memory.h>#define mod 1000000007#define N 1005#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)int dp[N][N][3][3],a[N];int main() {    int T,n,s,tmp;    int ans;    scanf("%d",&T);    while(T--) {        ans=0;        scanf("%d%d",&n,&s);        FOR(i,1,n)scanf("%d",&a[i]);        memset(dp,0,sizeof dp);        dp[1][a[1]][0][0]=1;        dp[1][a[1]][1][0]=1;        dp[1][0][0][0]=1;        dp[1][0][0][1]=1;        FOR(i,2,n)FOR(j,0,s) {            tmp=j+a[i];            if(tmp<=s)FOR(jj,0,2) {                dp[i][tmp][0][jj]=(dp[i][tmp][0][jj]+dp[i-1][j][0][jj])%mod;                dp[i][tmp][1][jj]=(dp[i][tmp][1][jj]+dp[i-1][j][0][jj])%mod;                dp[i][tmp][1][jj]=(dp[i][tmp][1][jj]+dp[i-1][j][1][jj])%mod;                dp[i][tmp][2][jj]=(dp[i][tmp][2][jj]+dp[i-1][j][1][jj])%mod;                dp[i][tmp][2][jj]=(dp[i][tmp][2][jj]+dp[i-1][j][2][jj])%mod;            }            FOR(ii,0,2) {                dp[i][j][ii][0]=(dp[i][j][ii][0]+dp[i-1][j][ii][0])%mod;                dp[i][j][ii][1]=(dp[i][j][ii][1]+dp[i-1][j][ii][0])%mod;                dp[i][j][ii][1]=(dp[i][j][ii][1]+dp[i-1][j][ii][1])%mod;                dp[i][j][ii][2]=(dp[i][j][ii][2]+dp[i-1][j][ii][1])%mod;                dp[i][j][ii][2]=(dp[i][j][ii][2]+dp[i-1][j][ii][2])%mod;            }        }        FOR(m,1,s)ans=(ans+dp[n][m][2][2])%mod;        printf("%d\n",(4LL*ans)%mod);    }    return 0;}