HDU 4301 Divide Chocolate （DP + 递推）

Divide Chocolate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1465    Accepted Submission(s): 682

Problem Description

It is well known that claire likes dessert very much, especially chocolate. But as a girl she also focuses on the intake of calories each day. To satisfy both of the two desires, claire makes a decision that each chocolate should be divided into several parts, and each time she will enjoy only one part of the chocolate. Obviously clever claire can easily accomplish the division, but she is curious about how many ways there are to divide the chocolate.

To simplify this problem, the chocolate can be seen as a rectangular contains n*2 grids (see above). And for a legal division plan, each part contains one or more grids that are connected. We say two grids are connected only if they share an edge with each other or they are both connected with a third grid that belongs to the same part. And please note, because of the amazing craft, each grid is different with others, so symmetrical division methods should be seen as different.

 

Input

First line of the input contains one integer indicates the number of test cases. For each case, there is a single line containing two integers n (1<=n<=1000) and k (1<=k<=2*n).n denotes the size of the chocolate and k denotes the number of parts claire wants to divide it into.

 

Output

For each case please print the answer (the number of different ways to divide the chocolate) module 100000007 in a single line.�

 

Sample Input

2 2 1 5 2

 

Sample Output

1 45

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

Recommend

zhuyuanchen520

 

题目的意思是:给你n*2这么大的巧克力 ,要分成M种，问你有多少种分法。

　　　　先定义dp[i][j][z],i表示第i列，j表示i列内分成j部分，z表示第i列的两块巧克力处于分开的状态还合并的。

所以 ， 如果添加第i+1列则有三种情况：

1,　　还是j部分，也就是不加部分 ：dp[i][j][1]=dp[i-1][j][0]*2+dp[i-1][j][1] ;   dp[i][j][0]=dp[i-1][j][0];

2,       加一部分，j+1：   dp[i][j][1]=dp[i-1][j-1][0]+dp[i-1][j-1][1];     dp[i][j][0]=dp[i-1][j-1][1]+dp[i-1][j-1][1]+2*dp[i-1][j-1][0];

3,       加两部分，j+2：   dp[i][j][0]=dp[i-1][j-2][1]+dp[i-1][j-2][0];

这里有个图可以帮助理解：http://blog.csdn.net/youngyangyang04/article/details/7764817

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=1010;
const int mod=100000007;

int dp[N][N+N][2];  //dp[i][j][z],i表示第i列，j表示i列内分成j部分，z表示第i列的两块巧克力处于分开的状态还合并的(0:分开,1:合并）

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,k;
    memset(dp,0,sizeof(dp));
    dp[1][1][1]=1;dp[1][2][0]=1;
    for(int i=2;i<N;i++){
        for(int j=1;j<=i*2;j++){
            dp[i][j][1]=(2*dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j-1][0]+dp[i-1][j-1][1])%mod;
            dp[i][j][0]=(dp[i-1][j][0]+2*dp[i-1][j-1][1]+2*dp[i-1][j-1][0]+dp[i-1][j-2][0]+dp[i-1][j-2][1])%mod;
        }
    }
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&k);
        printf("%d
",(dp[n][k][1]+dp[n][k][0])%mod);
    }
    return 0;
}