【HDU】1331 - Function Run Fun（记忆化递归）

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Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3922    Accepted Submission(s): 1918

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 12 2 210 4 650 50 50-1 7 18-1 -1 -1 

 

Sample Output

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

 

Source

Pacific Northwest 1999

 

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说白了就是动态规划的核心——记忆化搜索。

代码如下：

#include <cstdio>#include <stack>#include <queue>#include <cmath>#include <vector>#include <cstring>#include <algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long longLL num[22][22][22];LL solve(int a,int b,int c){if (a <= 0 || b <= 0 || c <= 0)return 1;else if (a > 20 || b > 20 || c > 20)return solve(20,20,20);else if (num[a][b][c] != -1)//记忆化搜索 return num[a][b][c];else if (a < b && b < c)return (num[a][b][c] = solve(a,b,c-1) + solve(a,b-1,c-1) - solve(a,b-1,c));elsereturn (num[a][b][c] = solve(a-1,b,c) + solve(a-1,b-1,c) + solve(a-1,b,c-1) - solve(a-1,b-1,c-1));}int main(){int a,b,c;while (~scanf ("%d %d %d",&a,&b,&c)){if (a == -1 && b == -1 && c == -1)break;CLR(num,-1);LL ans = solve(a,b,c);printf ("w(%d, %d, %d) = ",a,b,c);printf ("%lld\n",ans);}return 0;}