UVA 题目10128 Queue（DP）

最新推荐文章于 2019-10-30 00:21:52 发布

Jogging_Clown

最新推荐文章于 2019-10-30 00:21:52 发布

阅读量450

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/49903971

动态规划专栏收录该内容

194 篇文章

订阅专栏

探讨了n个不同高度的人排队的问题，从前边能看到m个人，从后边能看到r个人，使用动态规划方法解决有多少种可能的排列方式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There is a queue with N people. Every person has a different heigth. We can see P people, when we
are looking from the beginning, and R people, when we are looking from the end. Its because they
are having different height and they are covering each other. How many different permutations of our
queue has such a interesting feature?
Input
The input consists of T test cases. The number of them (1 ≤ T ≤ 10000) is given on the first line of
the input file.
Each test case begins with a line containing a single integer number N that indicates the number
of people in a queue (1 ≤ N ≤ 13). Then follows line containing two integers. The first integer
corresponds to the number of people, that we can see looking from the beginning. The second integer
corresponds to the number of people, that we can see looking from the end.
Output
For every test case your program has to determine one integer. Print how many permutations of N
people we can see exactly P people from the beginning, and R people, when we are looking from the
end.
Sample Input
3
10 4 4
11 3 1
3 1 2
Sample Output
90720
1026576

题目大意：n个高低不同的人排队，从前边看有m个人，从后边能看到r个人。问有多少种站法

思路：dp[i][j][k]表示i个人，从前边看j个，从后边看k个的排法种类数，把最矮的放在前边j+1（谁都挡不住），放在最后边k+1,（谁都挡不住），放在中间任意一个位置，j，k不变，于是递推公式dp[i][j][k]=dp[i-1][j-1][k]+dp[i-1][j][k-1]+(i*2)*dp[i-1][j][k]

ac代码

16452698

10128

Queue

Accepted

C++

0.003

2015-11-18 02:59:23

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
LL dp[22][22][22];
void fun()
{
    int i,j,k;
    dp[1][1][1]=1;
    for(i=2;i<=13;i++)
    {
        for(j=1;j<=i;j++)
        {
            int r=i-j+1;
            for(k=1;k<=r;k++)
            {
                dp[i][j][k]=dp[i-1][j][k-1]+dp[i-1][j-1][k]+(i-2)*dp[i-1][j][k];
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    fun();
    while(t--)
    {
        int n,m,r;
        scanf("%d%d%d",&n,&m,&r);
        printf("%lld\n",dp[n][m][r]);
    }
}