POJ_3046_Ant_Counting_(动态规划,多重集组合数)

描述

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http://poj.org/problem?id=3046

n种蚂蚁,第i种有ai个,不同种类的蚂蚁可以相互区分,但同一种类的蚂蚁不能相互区分,从这些蚂蚁中取出s,s+1,s+2,...,b-1,b个,问每种取的方式的取法数之和.

原型:多重集组合数:

n种物品,第i种有ai个.不同种类的物品可以相互区分,但同一种类的物品不能相互区分.从这些物品中取出m个,有多少种取法?

Ant Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4358		Accepted: 1689

Description

Bessie was poking around the ant hill one day watching the ants march to and fro while gathering food. She realized that many of the ants were siblings, indistinguishable from one another. She also realized the sometimes only one ant would go for food, sometimes a few, and sometimes all of them. This made for a large number of different sets of ants!

Being a bit mathematical, Bessie started wondering. Bessie noted
that the hive has T (1 <= T <= 1,000) families of ants which she
labeled 1..T (A ants altogether). Each family had some number Ni (1
<= Ni <= 100) of ants.

How many groups of sizes S, S+1, ..., B (1 <= S <= B <= A) can be formed?

While observing one group, the set of three ant families was seen as
{1, 1, 2, 2, 3}, though rarely in that order. The possible sets of
marching ants were:

3 sets with 1 ant: {1} {2} {3}

5 sets with 2 ants: {1,1} {1,2} {1,3} {2,2} {2,3}

5 sets with 3 ants: {1,1,2} {1,1,3} {1,2,2} {1,2,3} {2,2,3}

3 sets with 4 ants: {1,2,2,3} {1,1,2,2} {1,1,2,3}

1 set with 5 ants: {1,1,2,2,3}

Your job is to count the number of possible sets of ants given the data above.

Input

* Line 1: 4 space-separated integers: T, A, S, and B

* Lines 2..A+1: Each line contains a single integer that is an ant type present in the hive

Output

*
Line 1: The number of sets of size S..B (inclusive) that can be
created. A set like {1,2} is the same as the set {2,1} and should not be
double-counted. Print only the LAST SIX DIGITS of this number, with no
leading zeroes or spaces.

Sample Input

3 5 2 3
1
2
2
1
3

Sample Output

10

Hint

INPUT DETAILS:

Three types of ants (1..3); 5 ants altogether. How many sets of size 2 or size 3 can be made?

OUTPUT DETAILS:

5 sets of ants with two members; 5 more sets of ants with three members

Source

USACO 2005 November Silver

分析

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一.原型算法:

用dp[i][j]表示在前i种物品中取出j个的组合数.

那么可以从前(i-1)个中取(j-k)个,再从第i个中取k个,则有:

dp[i][j]=Σdp[i-1][j-k](0<=k<=min(a[i],j)).枚举i,j,k,这样的算法是O(n*m^2)的.

优化:

对Σdp[i-1][j-k](0<=k<=min(a[i],j))进行变形:

讨论a[i]与j的关系:

1.a[i]<j即min(a[i],j)=a[i]

则有:Σdp[i-1][j-k](0<=k<=min(a[i],j))=Σdp[i-1][j-1-k](0<=k<=min(a[i],j-1))+dp[i-1][j]-dp[i][j-1-a[i]].

即:dp[i][j]=dp[i-1][j]+dp[i][j-1]+dp[i-1][j-1-a[i]];

2.a[i]>=j即min(a[i],j)=j

则有:Σdp[i-1][j-k](0<=k<=min(a[i],j))=Σdp[i-1][(j-1)-k](0<=k<=min(a[i],j-1))+dp[i-1][j].

即:dp[i][j]=dp[i-1][j]+dp[i][j-1].

综上:

if(j--a[i])>= dp[i][j]=dp[i-][j]+dp[i][j-]-dp[i-][j--a[i]];

else dp[i][j]=dp[i-][j]+dp[i][j-];

继续优化,在空间上,dp只用到了i和i-1,可以考虑用滚动数组重复利用空间.

二.该题:

在原型的基础上最后进行一次统计,计算ans=Σ(dp[n][i])(s<=i<=t)即可.

 #include<cstdio>
 #include<algorithm>
 using namespace std;

 const int maxn=,maxm=*+,mod=1e6;
 int n,m,s,b;
 int a[maxn];
 int dp[][maxm];

 void solve()
 {
     dp[][]=dp[][]=;
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=m;j++)
         {
             if(j--a[i]>=)
             {
                 dp[i&][j]=(dp[i&][j-]+dp[(i-)&][j]-dp[(i-)&][j--a[i]]+mod)%mod;
             }
             else
             {
                 dp[i&][j]=(dp[i&][j-]+dp[(i-)&][j])%mod;
             }
         }
     }
     int ans=;
     for(int i=s;i<=b;i++)
     {
         ans=(ans+dp[n&][i])%mod;
     }
     printf("%d\n",ans);
 }

 void init()
 {
     scanf("%d%d%d%d",&n,&m,&s,&b);
     for(int i=;i<=m;i++)
     {
         int now;
         scanf("%d",&now);
         a[now]++;
     }
 }

 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("ant.in","r",stdin);
     freopen("ant.out","w",stdout);
 #endif
     init();
     solve();
 #ifndef ONLINE_JUDGE
     fclose(stdin);
     fclose(stdout);
     system("ant.out");
 #endif
     return ;
 }