AtCoder 2370 Piling

D - Piling Up

---

Time limit : 2sec / Memory limit : 256MB

Score : 900 points

Problem Statement

Joisino has a lot of red and blue bricks and a large box. She will build a tower of these bricks in the following manner.

First, she will pick a total of N bricks and put them into the box. Here, there may be any number of bricks of each color in the box, as long as there are N bricks in total. Particularly, there may be zero red bricks or zero blue bricks. Then, she will repeat an operation M times, which consists of the following three steps:

• Take out an arbitrary brick from the box.
• Put one red brick and one blue brick into the box.
• Take out another arbitrary brick from the box.

After the M operations, Joisino will build a tower by stacking the 2×M bricks removed from the box, in the order they are taken out. She is interested in the following question: how many different sequences of colors of these 2×M bricks are possible? Write a program to find the answer. Since it can be extremely large, print the count modulo 109+7.

Constraints

• 1≤N≤3000
• 1≤M≤3000

---

Input

Input is given from Standard Input in the following format:

N M

Output

Print the count of the different possible sequences of colors of 2×M bricks that will be stacked, modulo 109+7.

---

Sample Input 1

Copy

2 3

Sample Output 1

Copy

56

A total of six bricks will be removed from the box. The only impossible sequences of colors of these bricks are the ones where the colors of the second, third, fourth and fifth bricks are all the same. Therefore, there are 26−2×2×2=56 possible sequences of colors.

---

Sample Input 2

Copy

1000 10

Sample Output 2

Copy

1048576

---

Sample Input 3

Copy

1000 3000

Sample Output 3

Copy

693347555

题目大意：给一个盒子，盒子里面有n个砖头，这n个砖头的颜色可以自己确定，进行m次操作，每次的操作为：

1.  任意拿出一个砖头

2.  在盒子里放一个蓝色砖头和一个红色砖头

3.  再任意拿出一个砖头

现在给出n和m，求出有多少种拿出砖头的颜色的顺序。

思路：看到是方案数，很容易想到DP，我们设计状态为f[i][j]表示前i次操作，box里还剩j个红砖的方案数。哇！一眼秒！然后发现一些奇怪的问题：貌似会重复计数。我们发现，初始状态是自己设定的，所以可能导致看起来里面的数量不同，但是会重复计算。我们只能加一维限制条件，而一般的限制条件很明天会T，我们就只能加一维布尔型，这里表示为是否盒子里拿空过红砖。f[i][j][0/1]表示前i次操作，box里还剩j个红砖，是否拿空过红砖的方案数。然后就是一些简单转移。

这里给出一个几何理解：我们需要通过一些路径，使在平面直角坐标系中，从y轴走到x=2m处，我们不难看出，互相平行的折线是相同的情况，所以我们只要给折线标记一个上界或下界，就可以保证不重复计数。

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<cmath>#include<cstdlib>#include<ctime>using namespace std;const int MAXN=3010;const int Mod=1e9+7;int f[MAXN][MAXN][2];int main(){int n,m;scanf("%d%d",&n,&m);for (int i=0;i<=n;i++){f[0][i][0]=1;}for (int i=1;i<=m;i++){for (int j=0;j<=n;j++){int k=n-j;if (j){f[i][j][1]+=f[i-1][j][1];f[i][j][1]%=Mod;f[i][j][1]+=f[i-1][j-1][1];f[i][j][1]%=Mod;if (j==1){f[i][j][1]+=f[i-1][j][0];f[i][j][1]%=Mod;f[i][j][1]+=f[i-1][j-1][0];f[i][j][1]%=Mod;}else{f[i][j][0]+=f[i-1][j][0];f[i][j][0]%=Mod;f[i][j][0]+=f[i-1][j-1][0];f[i][j][0]%=Mod;}}if (k){f[i][j][1]+=f[i-1][j][1];f[i][j][1]%=Mod;f[i][j][1]+=f[i-1][j+1][1];f[i][j][1]%=Mod;if (j==0){f[i][j][1]+=f[i-1][j][0];f[i][j][1]%=Mod;f[i][j][1]+=f[i-1][j+1][0];f[i][j][1]%=Mod;}else{f[i][j][0]+=f[i-1][j][0];f[i][j][0]%=Mod;f[i][j][0]+=f[i-1][j+1][0];f[i][j][0]%=Mod;}}}}int ans=0;for (int i=0;i<=n;i++){ans+=f[m][i][1];if (ans>Mod){ans-=Mod;}}cout<<ans<<endl;    return 0;}