Pku1579 Function Run Fun （动态规划）递归函数

J - Function Run Fun

Time Limit:1000MS    Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

 

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

 

 

这本身就是一个递归函数，要是按照函数本身写递归式，结果肯定是TLE，这里我开了一个三维数组，从w(0,0,0)开始递推，逐步产生到w(20,20,20)的值，复杂度O(n^3).

总结：这道题是很地道的DP，因为它的子问题实在是太多了，所以将问题的结果保存起来。

 

#include<stdio.h> //0ms AC#include<string.h>int map[25][25][25]; //记忆化搜索int w(int a,int b,int c){if(a <= 0 || b <= 0 || c <= 0) return 1;if(a > 20 || b > 20 || c > 20) return w(20, 20, 20);if(map[a][b][c]!=-1) return map[a][b][c]; //递归中返回值if(a < b && b < c) map[a][b][c]=w(a,b,c-1) + w(a,b-1,c-1) - w(a,b-1,c);elsemap[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 map[a][b][c]; //最终递归返回值}int main(){int a,b,c;while(scanf("%d%d%d",&a,&b,&c)&&!(a==-1 && b==-1 && c==-1)){memset(map,-1,sizeof(map));printf("w(%d, %d, %d) = %d\n",a,b,c,w(a,b,c));}return 0;}