HDOJ题目2861 Stools（递推）

酒吧凳子排列问题

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探讨了一个涉及N个凳子、M个顾客及K个间隔的排列组合问题，通过动态规划算法求解特定条件下可能的排列数量。

Stools

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 626 Accepted Submission(s): 227

Problem Description

Patti and Terri run a bar in which there are 15 stools. One day, Darrell entered the bar and found that the situation how customers chose the stools were as follows:
OOEOOOOEEEOOOEO
O means that the stool in a certain position is used, while E means that the stool in a certain position is empty (here what we care is not who sits on the stool, but whether the stool is empty).As the example we show above, we can say the situation how the 15 stools is used determines 7 intervals (as following):
OO E OOOO EEE OOO E O

Now we postulate that there are N stools and M customers, which make up K intervals. How many arrangements do you think will satisfy the condition?

Input

There are multi test cases and for each test case:
Each case contains three integers N (0<N<=200), M (M<=N), K (K<=20).

Output

For each test case print the number of arrangements as described above. (All answers is fit in 64-bit.)

Sample Input

  
   3 1 3
4 2 4

Sample Output

1
2

Source

2009 Multi-University Training Contest 5 - Host by NUDT

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a[n][m][k]=a[n-1][m][k]+b[n-1][m][k-1];
b[n][m][k]=a[n-1][m-1][k-1]+b[n-1][m-1][k];
ans[n][m][k]=a[n][m][k]+b[n][m][k]。

a[n][m][k]：n个座位、m个人、k段、最后一个位置么有人，的情况数；

b[n][m][k]：n个座位、m个人、k段、最后一个位置有人，的情况数。

ac代码

#include<stdio.h>
#include<string.h>
__int64 a[220][220][22],b[220][220][22];
void init()
{
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	a[1][0][1]=1;
	b[1][1][1]=1;
	int i,j,k;
	for(i=2;i<=200;i++)
	{
		for(j=0;j<=i;j++)
		{
			for(k=1;k<=i&&k<22;k++)
			{
				a[i][j][k]=a[i-1][j][k]+b[i-1][j][k-1];
				if(j)
					b[i][j][k]=a[i-1][j-1][k-1]+b[i-1][j-1][k];
			}
		}
	}
}
int main()
{
	init();
	int n,m,k;
	while(scanf("%d%d%d",&n,&m,&k)!=EOF)
	{
		printf("%I64d\n",a[n][m][k]+b[n][m][k]);
	}
}