1579、Function Run Fun(记忆化搜索)

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21964		Accepted: 10914

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

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long;

using namespace std;

int A,B,C;
int dp[25][25][25];
const int INF=0x3f3f3f3f;

int dfs(int a,int b,int c){   //记忆化搜索
    if(a<=0||b<=0||c<=0) return 1;
    if(a>20||b>20||c>20) return dfs(20,20,20);
    if(dp[a][b][c]!=INF) return dp[a][b][c];     //记忆化搜索  之前搜到过直接返回
    if(a<b&&b<c) dp[a][b][c]=dfs(a,b,c-1)+dfs(a,b-1,c-1)-dfs(a,b-1,c);
    else dp[a][b][c]=dfs(a-1,b,c)+dfs(a-1,b-1,c)+dfs(a-1,b,c-1)-dfs(a-1,b-1,c-1);
    return dp[a][b][c];

}

int main(){
    ios::sync_with_stdio(false);
    while(cin>>A>>B>>C&&(A!=-1||B!=-1||C!=-1)){
        memset(dp,INF,sizeof(dp));
        printf("w(%d, %d, %d) = %d
",A,B,C,dfs(A,B,C));
    }
    return 0;
}