POJ 1579 Function Run Fun 记忆化搜索

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17499 Accepted: 8963

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 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

Source

Pacific Northwest 1999

[Submit]   [Go Back]   [Status]   [Discuss]

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 100005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;int a,b,c;int cnt=1;int dp[30][30][30];int fun(int a,int b,int c){    if(a<=0||b<=0||c<=0)return 1;    if(a>20||b>20||c>20)        return dp[20][20][20]=fun(20,20,20);    if(dp[a][b][c])return dp[a][b][c];    if(a<b&&b<c)        return dp[a][b][c]=fun(a,b,c-1)+fun(a,b-1,c-1)-fun(a,b-1,c);    return dp[a][b][c]=fun(a-1,b,c)+fun(a-1,b-1,c)+fun(a-1,b,c-1)-fun(a-1,b-1,c-1);}int main(){    while(scanf("%d %d %d",&a,&b,&c)!=EOF){        if(a==-1&&b==-1&&c==-1)break;   MT(dp,0);        printf("w(%d, %d, %d) = %d\n",a,b,c,fun(a,b,c));    }    return 0;}/*1 1 12 2 210 4 650 50 50-1 7 18-1 -1 -1*/