HDU 4489 The King’s Ups and Downs dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4489

The King’s Ups and Downs

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) #### 问题描述 > The king has guards of all different heights. Rather than line them up in increasing or decreasing height order, he wants to line them up so each guard is either shorter than the guards next to him or taller than the guards next to him (so the heights go up and down along the line). For example, seven guards of heights 160, 162, 164, 166, 168, 170 and 172 cm. could be arranged as: > or perhaps: > The king wants to know how many guards he needs so he can have a different up and down order at each changing of the guard for rest of his reign. To be able to do this, he needs to know for a given number of guards, n, how many different up and down orders there are: > > For example, if there are four guards: 1, 2, 3,4 can be arrange as: > > 1324, 2143, 3142, 2314, 3412, 4231, 4132, 2413, 3241, 1423 > > For this problem, you will write a program that takes as input a positive integer n, the number of guards and returns the number of up and down orders for n guards of differing heights. #### 输入 > The first line of input contains a single integer P, (1 <= P <= 1000), which is the number of data sets that follow. Each data set consists of single line of input containing two integers. The first integer, D is the data set number. The second integer, n (1 <= n <= 20), is the number of guards of differing heights. #### 输出 > For each data set there is one line of output. It contains the data set number (D) followed by a single space, followed by the number of up and down orders for the n guards.

样例输入

4
1 1
2 3
3 4
4 20

样例输出

1 1
2 4
3 10
4 740742376475050

题意

给你n个数（1到n），问排出高低高，，，或低高低，，，的所有情况数。

题解

考虑最后一个数n摆放的位置，枚举在它左边放x个数，右边放n-1-x个数，然后转移。
需要用到一个结论：对于任意的n>=2，高低高和低高低的情况数都是相等的！

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=22;

LL dp[maxn],C[maxn][maxn];

void get_C(){
    C[0][0]=1;
    for(int i=1;i<maxn;i++){
        C[i][0]=1;
        for(int j=1;j<=i;j++){
            C[i][j]=C[i-1][j-1]+C[i-1][j];
        }
    }
}

void pre(){
    get_C();
    dp[0]=dp[1]=1;
    dp[2]=2;
    for(int i=3;i<=20;i++){
        for(int x=0;x<i;x++){
            LL lef=x<=1?dp[x]:dp[x]/2;
            int y=i-x-1;
            LL rig=y<=1?dp[y]:dp[y]/2;
            dp[i]+=C[i-1][x]*lef*rig;
        }
    }
}

int main() {
    pre();
    int tc;
    scanf("%d",&tc);
    while(tc--){
        int kase,n;
        scf("%d%d",&kase,&n);
        prf("%d %lld
",kase,dp[n]);
    }
    return 0;
}

//end-----------------------------------------------------------------------