HDU 1331 Function Run Fun（记忆化搜索）

 

 

          Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1387    Accepted Submission(s): 709

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

 

 

Source

Pacific Northwest 1999

 

 

Recommend

Ignatius.L

 

 

题意：给你一些递归条件，然后告诉你a,b,c的值,要求算出调用w(a,b,c)函数后所返回的值。
分析：一道典型的自底向上的记忆化搜索问题。经典地道的DP动态规划问题。如果直接用递归做肯定会超时，因为它的子问题实在是太多了，递归过程有许多重复计算的部分，解决办法就是用记忆化搜索解决，自底向上的递推，将已求解出来的结果存到数组里保存下来，下次再递归到此处直接读取答案就行，避免重复计算。复杂度O(n^3)。

 

 

#include<iostream>
using namespace std;
int x[25][25][25]={0};
int w(int a,int b,int c)
{
    if(a<=0||b<=0||c<=0)
        return 1;
    if(x[a][b][c]!=0));//关键语句,避免重复计算
        return x[a][b][c];
    if(a<b&&b<c)
    {
        x[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
        return x[a][b][c];
    }
    else 
    {
        x[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);
        return x[a][b][c];
    }
}
int main()
{
    int a,b,c;
    while(~scanf("%d %d %d",&a,&b,&c))
    {
        if(a==-1&&b==-1&&c==-1)
            break;
        if(a<=0||b<=0||c<=0)
        {
            printf("w(%d, %d, %d) = %d\n",a,b,c,1);
            continue;
        }
        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;
}