记忆化搜索:POJ1579-Function Run Fun(最基础的记忆化搜索)
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Function Run Fun
Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 14815Accepted: 7659
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.
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
解题心得:
1、这是一个很经典的记忆化搜索,算是记忆化搜索之中最简单的。其实记忆化搜索就是dp,从小到大以此往上转移。就这个题本身来说是一个很很经典的从下到上的递推。
2、题目描述也很清楚,下一个的结果要从上一个的结果中的出,如果下一个的结果未知,那么就会产生大量的递归然后复杂度爆炸,由前到后依次得到答案,上一个答案从下一个中得出,就是动态规划一个很明显的特征。
#include<cstring>#include<stdio.h>const int maxn = 25;struct num{ bool flag[maxn][maxn][maxn]; int Num[maxn][maxn][maxn];} N;//记录出现过的结果int w(int a,int b,int c){ //根据题意就好 if(a <= 0 || b <= 0 || c <= 0) return 1; if(a > 20 || b > 20 || c > 20) return w(20,20,20); if(N.flag[a][b][c]) return N.Num[a][b][c]; if(a<b && b<c) return w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c); return 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; //预处理一下 for(int i=1; i<=20; i++) for(int j=1; j<=20; j++) for(int k=1; k<=20; k++) { a = w(i,j,k); N.flag[i][j][k] = true; N.Num[i][j][k] = a; } while(scanf("%d%d%d",&a,&b,&c)) { if(a == -1 && b == -1 && c == -1) break; int ans = w(a,b,c); printf("w(%d, %d, %d) = %d\n",a,b,c,ans); }
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