POJ 3046 Ant Counting(多重集组合数）（基础组合dp)

Ant Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6467		Accepted: 2397

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

dp[i][j]表示从i个家族中取j只蚂蚁，m=min(j,a[i]);可以从第i个家族中取k个（0<=k<=m）
则从前i-1个家族中取j-k个因此dp[i][j]=∑dp[i-1][j-k](0<=k<=m);
按照此种方法求解时间复杂度O(100*T*A);
优化：
（1）.当j<=a[i],即j-1<a[i]时：m=j;此时
dp[i][j]=dp[i-1][0]+dp[i-1][1]+……+dp[i-1][j-1]+dp[i-1][j];
而dp[i-1][0]+dp[i-1][1]+….+dp[i-1][j-1]=dp[i][j-1];因此得出
dp[i][j]=dp[i][j-1]+dp[i-1][j];
(2).当j>a[i]时，m=a[i];此时
dp[i][j]=dp[i-1][j-a[i]]+dp[i-1][j-a[i]+1]+….+dp[i-1][j];
类似以上方法，可得
dp[i-1][j-a[i]-1]+dp[i-1][j-a[i]]+…+dp[i-1][j-1]=dp[i][j-1];由此得出
dp[i][j]=dp[i][j-1]-dp[i-1][j-a[i]-1]+dp[i-1][j]。
这样转化以后，我们在求每一个状态时可以直接由之前状态得到，
这样时间复杂度就降到了O(T*A)。

O(100*T*A)代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
int a[1002],mod=1e6;
int dp[2][100002];//滚动数组节省内存
using namespace std;
int main()
{
int T,A,S,B;
while(scanf("%d%d%d%d",&T,&A,&S,&B)!=EOF)
{
memset(a,0,sizeof(a));
for(int i=1;i<=A;i++)
{
int t;scanf("%d",&t);
a[t]++;
}
memset(dp,0,sizeof(dp));
dp[0][0]=1,dp[1][0]=1;//从每个家族取零个都有一种方法。
for(int i=1;i<=T;i++)
{
for(int j=1;j<=B;j++)//B之后的状态不必考虑
{
int m=min(j,a[i]);
dp[i%2][j]=0;//由于开的是滚动数组，dp[i%2][j]在之前可能已经被赋值。

for(int k=0;k<=m;k++)
{
dp[i%2][j]=(dp[i%2][j]+dp[(i-1)%2][j-k])%mod;
}
}
}
int s=0;
for(int i=S;i<=B;i++)
s=(s+dp[T%2][i])%mod;
printf("%d\n",s);
}
return 0;
}

O(T*A)代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
int a[1003], dp[2][100002],mod=1e6;
using namespace std;
int main()
{
int T, A, S, B;
while (scanf("%d%d%d%d", &T, &A, &S, &B) != EOF)
{
memset(a, 0, sizeof(a));
for (int i = 1; i <= A; i++)
{
int t; scanf("%d", &t);
a[t]++;
}
memset(dp, 0, sizeof(dp));
dp[0][0] = 1; dp[1][0] = 1;
int sum = 0;
for (int i = 1; i <= T; i++)
{
for (int j = 1; j <=B; j++)
{
if (j <= a[i])
dp[i&1][j] = (dp[i&1][j - 1] + dp[(i - 1)&1][j])%mod;
else //为了避免相减出现负数的情况，每次加mod
dp[i&1][j] = (dp[i&1][j - 1] + dp[(i - 1)&1][j] - dp[(i-1)&1][j - 1 - a[i]]+mod)%mod;
}
}
for(int i=S;i<=B;i++)
sum=(sum+dp[T&1][i])%mod;
printf("%d\n", sum%mod);
}
return 0;
}