poj 1579 Function Run Fun
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.
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.
题意很好理解,不解释
数组还函数的名字不要起一样的,要大小写区分
分析,由于子问题过多,每次递归的时候就记录结果
#include<stdio.h> #include<string.h> const int MAXN=25; int w[MAXN][MAXN][MAXN]; int W(int a,int b,int c) { if(w[a][b][c]) return w[a][b][c]; if(a<=0 || b<=0 || c<=0) return w[a][b][c]=1; if(a<b && b<c) return w[a][b][c]=W(a,b,c-1)+W(a,b-1,c-1)-W(a,b-1,c); else return w[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); } int main() { int a,b,c; while(scanf("%d%d%d",&a,&b,&c)) { memset(w,0,sizeof(w)); if(a==-1 && b==-1 && c==-1) break; if(a<=0 || b<=0 || c<=0) printf("w(%d, %d, %d) = 1\n",a,b,c); else if(a>20 || b>20 || c>20) printf("w(%d, %d, %d) = %d\n",a,b,c,W(20,20,20)); else printf("w(%d, %d, %d) = %d\n",a,b,c,W(a,b,c)); } return 0; }
从别人那学来的
预处理
void init() { for(a=0;a<=20;a++) for(b=0;b<=20;b++) for(c=0;c<=20;c++) m[a][b][c]=1; for(a=1;a<=20;a++) for(b=1;b<=20;b++) for(c=1;c<=20;c++){ if(a < b && b < c) m[a][b][c]=m[a][b][c-1]+m[a][b-1][c-1]-m[a][b-1][c]; else m[a][b][c]=m[a-1][b][c]+m[a-1][b-1][c]+m[a-1][b][c-1]-m[a-1][b-1][c-1]; } }