E - Camels
Crawling in process...Crawling failedTime Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 14E
Description
Bob likes to draw camels: with a single hump, two humps, three humps, etc. He draws a camel by connecting points on a coordinate plane. Now he's drawing camels with t humps, representing them as polylines in the plane. Each polyline consists of n vertices with coordinates (x1, y1), (x2, y2), ..., (xn, yn). The first vertex has a coordinate x1 = 1, the second — x2 = 2, etc. Coordinates yi
- there should be t humps precisely, i.e. such indexes j (2 ≤ j ≤ n - 1), so that yj - 1 < yj > yj + 1,
- there should be precisely t - 1 such indexes j (2 ≤ j ≤ n - 1), so that yj - 1 > yj < yj + 1,
- no segment of a polyline should be parallel to the Ox-axis,
- all yi
For a series of his drawings of camels with t humps Bob wants to buy a notebook, but he doesn't know how many pages he will need. Output the amount of different polylines that can be drawn to represent camels with t humps for a given number n.
Input
The first line contains a pair of integers n and t (3 ≤ n ≤ 20, 1 ≤ t ≤ 10).
Output
Output the required amount of camels with t
Sample Input
Input
6 1
Output
6
Input
4 2
Output
0
Sample Output
Hint
In the first sample test sequences of y-coordinates for six camels are: 123421, 123431, 123432, 124321, 134321 и 234321 (each digit corresponds to one value of yi).
思路：四维DP，d[n][t][j][s] n代表点数（步长），t代表尖峰数，j代表高度，s代表上升（0）还是下降（1）。
E(dp[1][1][j][0])=1,用上升峰代表完成一个尖峰。初始值看做一个上升峰。
初始值看成上升峰，那么步长为2就会出现下降峰，但这是不合法的，把它都置0就OK了。
f[n][t][i][0]+=f[n-1][t-1][j][1]+f[n-1][t][j][0];///更新上升峰（1：上升峰再加一个上升，2：下降峰加个上升）
f[n][t][i][1]+=f[n-1][t][j][1]+f[n-1][t][j][0];///更新下降峰（1：下降峰再加一个下降，2：上升峰加个下降）
失误：比赛的时候想过用DP，但觉得搜索+减枝也行，就直接搜了，后来才发现搜索会爆。。。
#include<iostream>
#include<cstring>
using namespace std;
const int mm=22;
int f[mm][mm][mm][2];
int main()
{
memset(f,0,sizeof(f));
for(int i=1;i<5;i++)
f[1][1][i][0]=1;///边界上升尖峰为1
///0代表上升，1下降,上升时尖峰减1
for(int k=2;k<=20;k++)
for(int l=1;l<=10;l++)
for(int i=1;i<=4;i++)
{
for(int j=1;j<i;j++)///更新上升峰（1：上升峰再加一个上升，2：下降峰加个上升）
f[k][l][i][0]+=f[k-1][l-1][j][1]+f[k-1][l][j][0];
if(k==2)f[k][l][i][1]=0;///2步不可能的情况
else for(int j=i+1;j<5;j++)///更新下降峰（1：下降峰再加一个下降，2：上升峰加个下降）
f[k][l][i][1]+=f[k-1][l][j][1]+f[k-1][l][j][0];
}
int n,t;
int ans;
while(cin>>n>>t)
{ ans=0;
for(int i=1;i<=4;i++)
ans+=f[n][t][i][1];
cout<<ans<<"\n";
}
}
