POJ 1579 Function Run Fun【记忆化搜索】
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递归和打表找规律或者记忆化搜索
题目链接:http://poj.org/problem?id=1579
2012年暑假组队后第一场个人训练赛
Function Run Fun
Time Limit:1000MS
Memory Limit:10000K
Total Submissions:13148
Accepted:6836
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
Pacific Northwest 1999
上午比赛的继续http://172.16.24.47:33/judge/contest/view.action?cid=5#problem/A
我的思路,打表找规律。
常规思路:
记忆化搜索(把计算的结果存在数组中,避免重复计算,从而避免超时)
对于这些题目,还是常规思路的好
我的代码:
//AC 132k 16ms(by C++) POJ 1579#include<stdio.h>#include<math.h>int w(int a,int b,int c){if(a<=0 || b<=0 || c<=0)return 1;else if(a>20 || b>20 || c>20)return 1048576; // 只要有一个数大于20,则返回w(20,20,20)=1048576,即2的20次方else if(b>=a || c>=a) //只要后面两个数有一个大于a,则返回2的a次方(开始超时后,打表得出的规律。。。)return (double)pow((double)2,a); //注意pow函数的强制类型转换,一不小心就编译错误了else if(a<b && b<c)return w(a,b,c-1)+w(a,b-1,c)-w(a,b-1,c);elsereturn 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;double ans;while(scanf("%d%d%d",&a,&b,&c)!=EOF){if(a==-1 && b==-1 && c==-1)break;else{printf("w(%d, %d, %d) = ",a,b,c);ans=w(a,b,c);}printf("%.0lf\n",ans);}return 0;}
我同学的常规思路的代码:
//叶//思路:运用递归会超时,故而只要计算出前w[i][j][k](0<=k<=20&&0<=i<=20&&0<=j<=20) #include<stdio.h>#include<iostream>using namespace std;int w[21][21][21];void build(){int i,j,k;for(i=0;i<=20;i++)for(j=0;j<=20;j++)for(k=0;k<=20;k++){if(i<=0 || j<=0 || k<=0)w[i][j][k]=1;else if(i<j && j<k)w[i][j][k]=w[i][j][k-1]+w[i][j-1][k-1]-w[i][j-1][k];elsew[i][j][k]=w[i-1][j][k]+w[i-1][j-1][k]+w[i-1][j][k-1]-w[i-1][j-1][k-1] ;}}int main(){ build();int a,b,c,s;while(scanf("%d%d%d",&a,&b,&c)!=EOF && !(a==-1 && b==-1 && c==-1)){ if(a<=0||b<=0|c<=0)s=1;else if(a>20||b>20||c>20)s=w[20][20][20];elses=w[a][b][c]; printf("w(%d, %d, %d) = %d\n",a,b,c,s);}return 0;}
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