poj1579

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12713		Accepted: 6621

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 

解析

　　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).

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．这个例子就非常典型。总体来说这个题目还是非常简单的，不过这个思想是地道的动态规划。

参考代码

#include<stdio.h>

#define max_n 21
int main(){
    int ava[max_n][max_n][max_n];
    int i,j,k;
    int a,b,c;
    for(i=0;i<=20;i++)
for(j=0;j<=20;j++){
ava[i][j][0]=1;
ava[i][0][j]=1;
ava[0][i][j]=1;
}
for(i=1;i<=20;i++)
for(j=1;j<=20;j++)
for(k=1;k<=20;k++){
if(i<j&&j<k)
ava[i][j][k]=ava[i][j][k-1]+ava[i][j-1][k-1]-ava[i][j-1][k];
else
ava[i][j][k]=ava[i-1][j][k]+ava[i-1][j-1][k]+ava[i-1][j][k-1]-ava[i-1][j-1][k-1];
}
while(true){
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);
else if(a>=20||b>=20||c>=20)printf("w(%d, %d, %d) = %d\n",a,b,c,ava[20][20][20]);
else
printf("w(%d, %d, %d) = %d\n",a,b,c,ava[a][b][c]);
}
return 0;
}