zoukankan      html  css  js  c++  java
  • 01背包-第k优解

    The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: 

    Here is the link: http://acm.hdu.edu.cn/showproblem.php?pid=2602 

    Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum. 

    If the total number of different values is less than K,just ouput 0.

    Input

    The first line contain a integer T , the number of cases. 
    Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need.

    And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone. 

    Output

    One integer per line representing the K-th maximum of the total value (this number will be less than 2 31). 
    Sample Input

    3
    5 10 2
    1 2 3 4 5
    5 4 3 2 1
    5 10 12
    1 2 3 4 5
    5 4 3 2 1
    5 10 16
    1 2 3 4 5
    5 4 3 2 1

    Sample Output

    12
    2
    0
    #include <iostream>  
    #include <cstdio>  
    using namespace std;  
    #define max(a,b)    ((a)>(b)?(a):(b))  
    const int maxn = 1005;  
    int main()  
    {  
        int T;  
        scanf("%d", &T);  
        int dp[maxn][33], val[maxn], vol[maxn], A[33], B[33];  
        while (T--)  
        {  
            int n, v, k;  
            scanf("%d %d %d", &n, &v, &k);  
            int i, j, kk;  
            for (i=0; i<n; i++) scanf("%d", &val[i]);  
            for (i=0; i<n; i++) scanf("%d", &vol[i]);  
            memset(dp, 0, sizeof(dp));  
      
            int a, b, c;  
            for (i=0; i<n; i++)  
                for (j=v; j>=vol[i]; j--)  
                {  
                    for (kk=1; kk<=k; kk++)  
                    {  
                        A[kk] = dp[j-vol[i]][kk] + val[i];  
                        B[kk] = dp[j][kk];  
                    }  
                    A[kk] = -1, B[kk] = -1;  
                    a = b = c = 1;  
                    while (c<=k && (A[a] != -1 || B[b] != -1))  
                    {  
                        if (A[a] > B[b])  
                            dp[j][c] = A[a++];  
                        else  
                            dp[j][c] = B[b++];  
                        if (dp[j][c] != dp[j][c-1])  
                            c++;  
                    }  
                }  
      
            printf("%d
    ", dp[v][k]);  
        }  
        return 0;  
    }
  • 相关阅读:
    多重网格法简介(Multi Grid)
    数值分析方法库
    离散外微积分(DEC:Discrete Exterior Calculus)基础
    楔积(Wedge Procut)
    牛顿迭代法(Newton's Method)
    四元素还是向量?
    曲率(Curvature)
    共变导数(Covariant Derivative)
    正定矩阵(Positive-definite Matrix)
    散度(Divergence)和旋度(Curl)
  • 原文地址:https://www.cnblogs.com/carry-2017/p/7367544.html
Copyright © 2011-2022 走看看