【CF Gym 100637K】 Microcircuits 【DP】【环上n个点，问不相交的连k条线的方案数】

传送门：

描述：

K. Microcircuits

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You probably know how microcircuits look like. First of all it is important to pay special attention to connections. Contacts on the circuit are connected by lines. If two lines do not intersect then there is no connection between
respective contacts.

You are holding a circular circuit with n contacts around the borderline. You have to calculate the number of possibilities
to put exactly knon-intersecting lines each connecting two contacts.

Input

Single line contains two integers n and k (1 ≤ k ≤ n ≤ 40).

Output

Output the required number.

Examples

input

4 2

output

2

input

4 3

output

0

题意：

给你n个点，将这些点放在一个环上，问你不相交的连k条线的方案数。（没有重点）

思路：

dp[i][j]表示i个点连j条线的方案数，那么新加一个点i，

情况1：i没有和之前的点相连，方案数为dp[i-1][j]；

情况2：i和p号点相连(0<p<j)，那么就i-p这条线把所有点划分成两部分，然后枚举一些子问题的连边数，方案数为sum( dp[p-1][q]*dp[i-1-p][j-1-q] ) (0<=q<j)

复杂度：

O(n^4)

代码：

#include <bits/stdc++.h>
using namespace std;

#define ll __int64
#define rep(i,k,n) for(int i=k;i<=n;i++)

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

ll dp[50][50];

int main(){
int k,n;
read(n),read(k);
rep(i, 0, n)dp[i][0] = 1;
rep(i, 1, n){
rep(j, 0, k){
dp[i][j] = dp[i-1][j];
rep(p, 1, i-1){
rep(q, 0, j-1){
dp[i][j] += dp[p-1][q] * dp[i-1-p][j-1-q];
}
}
}
}
print(dp
[k]);
return 0;
}