poj 1579(动态规划初探之记忆化搜索)

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17843		Accepted: 9112

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

记忆化搜索。照着思路走。不过把已经搜索过的就不要重复计算了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include <string.h>
#include <math.h>
using namespace std;

int f[21][21][21];
int dfs(int a,int b,int c){
    if(a<=0||b<=0||c<=0) return 1;
    if(a>20||b>20||c>20) return dfs(20,20,20);
    if(f[a][b][c]!=-1) return f[a][b][c];
    if(a<b&&b<c) f[a][b][c] = dfs(a,b,c-1)+dfs(a,b-1,c-1)-
        dfs(a,b-1,c);
    else f[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 f[a][b][c];
}
int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF){
        if(a==-1&&b==-1&&c==-1) break;
        memset(f,-1,sizeof(f));
        int ans = dfs(a,b,c);
        printf("w(%d, %d, %d) = %d
",a,b,c,ans);
    }
    return 0;
}