POJ 1579-Function Run Fun（记忆化搜索-递归）

Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18930 Accepted: 9612

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

题目意思：

给出a,b,c，按以下递归方式计算w(a, b, c)：

①if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:1；
②if a > 20 or b > 20 or c > 20, then w(a, b, c) returns:w(20, 20, 20)
③if 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)
④otherwise 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)

解题思路：

直接递归必然超时，所以一定要记忆化搜索。

用dp[a][b][c]记录w(a,b,c)的值，每次判断一下如果计算过就不再重复计算，注意防止下标越界的问题，要先处理负数再处理大数。

#include <iostream>#include <cstdio>#include <cstring>#include <bitset>#include <map>#include <iomanip>#include <algorithm>#define MAXN 21#define INF 0xfffffffusing namespace std;/*if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns:1if 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)*/int dp[MAXN][MAXN][MAXN];int w(int a,int b,int c){    if(a<=0||b<=0||c<=0) return 1;//为了防止下标越界(为负)所以先处理    if(a>20||b>20||c>20) return dp[20][20][20]=w(20,20,20);//为了防止下标越界(大于20)所以先处理    if(dp[a][b][c]!=0) return dp[a][b][c];//已经计算过的不再重复计算    if(a<b&&b<c)  return dp[a][b][c]=(w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c));    return dp[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(){#ifdef ONLINE_JUDGE#else    freopen("G:/cbx/read.txt","r",stdin);    //freopen("G:/cbx/out.txt","w",stdout);#endif    ios::sync_with_stdio(false);    cin.tie(0);    memset(dp,0,sizeof(dp));    int a,b,c;    while(cin>>a>>b>>c)    {        if(a==-1&&b==-1&&c==-1) break;        cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<w(a,b,c)<<endl;    }    return 0;}