Function Run Fun
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1475 Accepted Submission(s): 752
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
Recommend
Ignatius.L
思路：水题一道，记忆化搜索
#include<iostream>
#include<cstring>
using namespace std;
const int mm=24;
int w[mm][mm][mm];
int dfs(int a,int b,int c)
{
if(a<=0||b<=0||c<=0)return 1;
if(a>20||b>20||c>20)return dfs(20,20,20);
if(w[a][b][c])return w[a][b][c];
if(a<b&&b<c)w[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
else w[a][b][c]=dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1);
return w[a][b][c];
}
int main()
{
memset(w,0,sizeof(w));
int a,b,c;
while(cin>>a>>b>>c)
{
if(a==-1&&b==-1&&c==-1)break;
cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<dfs(a,b,c)<<"\n";
}
}