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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/liyinggang/p/5369904.html
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