poj 1579 Function Run Fun(记忆化搜索)
2016-08-31 16:32
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Function Run Fun
DescriptionWe 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. InputThe 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.OutputPrint the value for w(a,b,c) for each triple.Sample Input
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 18256 | Accepted: 9328 |
1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1Sample 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
思路:递归+通过记录消除冗余
代码:
#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int issolved[100][100][100];int solve(int a,int b,int c){if(a<=0||b<=0||c<=0) return 1;if(a>20||b>20||c>20) return solve(20,20,20);if(issolved[a][b][c]) return issolved[a][b][c];if(a<b&&b<c) return issolved[a][b][c]=solve(a,b,c-1)+solve(a,b-1,c-1)-solve(a,b-1,c);else return issolved[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 x,y,z;memset(issolved,0,sizeof(issolved));while(~scanf("%d%d%d",&x,&y,&z)){if(x==-1&&y==-1&&z==-1) break;int ans=solve(x,y,z);printf("w(%d, %d, %d) = %d\n",x,y,z,ans);}return 0;}
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