记忆化搜索 dp学习~2

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1331

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3459    Accepted Submission(s): 1707

Problem Description

We all love recursion! Don't we?

Consider a three-parameter recursive function w(a, b, c):

if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:
1

if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:
w(20, 20, 20)

if a < b and b < c, then w(a, b, c) returns:
w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c)

otherwise it returns:
w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1)

This is an easy function to implement. The problem is, if implemented directly, for moderate values of a, b and c (for example, a = 15, b = 15, c = 15), the program takes hours to run because of the massive recursion.

 

Input

The input for your program will be a series of integer triples, one per line, until the end-of-file flag of -1 -1 -1. Using the above technique, you are to calculate w(a, b, c) efficiently and print the result.

 

Output

Print the value for w(a,b,c) for each triple.

 

Sample Input

1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1

 

Sample Output

w(1, 1, 1) = 2 w(2, 2, 2) = 4 w(10, 4, 6) = 523 w(50, 50, 50) = 1048576 w(-1, 7, 18) = 1

dp学习~2

动态规划的两种使用动机：

•自底向上的递推。

•利用递归时产生大量重叠子问题，进行记忆化求解。

其中记忆化搜索：

•形式上是搜索，但是把搜索到的一些解用动态规划的思想和模式保存下来。

•特点：

•1.一般来说dp总是需要遍历所有状态，搜索却不需要。

•2.搜索可以剪枝，可能还会剪去大量不必要状态。

•3.可能会比较容易编写，且效率不错。

这个题在使用递归函数的时候很明显会重复计算很多的状态，所以这里用记忆化搜索，思想就是算过的状态直接返回即可

特别注意一个问题就是，在算出一个问题的解以后一定要赋值给对应的dp[i][j][k];——更新dp数组

代码;

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 21;
int dp[N][N][N];

void init()
{
    memset(dp,-1,sizeof(dp));
    for(int i = 0; i <= 21; i++)
    {
        for(int j = 0; j <=21; j++)
        {
            dp[0][i][j] = dp[i][j][0] = dp[i][0][j] = 1;
        }
    }
}
int w(int a, int b, int c)
{
    if(dp[a][b][c]!=-1) return dp[a][b][c];
    if(a<b&&b<c){
        return dp[a][b][c] = w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
    }
    else return dp[a][b][c] = w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1);
}
int main()
{
    init();
    int a, b, c;
    while(~scanf("%d%d%d",&a,&b,&c))
    {
        //printf("w = %d\n",w(20,20,20));
        if(a==-1&&b==-1&&c==-1) return 0;
        else if(a<=0||b<=0||c<=0) printf("w(%d, %d, %d) = 1\n",a,b,c);
        else if(a>20||b>20||c>20) printf("w(%d, %d, %d) = %d\n",a,b,c,w(20,20,20));
        else printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));
    }
    return 0;
}