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  • 2015 多校联赛 ——HDU5303(贪心)

    Delicious Apples

    Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
    Total Submission(s): 1371    Accepted Submission(s): 448

    Problem Description
    There are n apple trees planted along a cyclic road, which is L metres long. Your storehouse is built at position 0 on that cyclic road.
    The ith tree is planted at position xi, clockwise from position 0. There are ai delicious apple(s) on the ith tree.

    You only have a basket which can contain at most K apple(s). You are to start from your storehouse, pick all the apples and carry them back to your storehouse using your basket. What is your minimum distance travelled?

    1n,k105,ai1,a1+a2+...+an105
    1L109
    0x[i]L

    There are less than 20 huge testcases, and less than 500 small testcases.
     
    Input
    First line: t, the number of testcases.
    Then t testcases follow. In each testcase:
    First line contains three integers, L,n,K.
    Next n lines, each line contains xi,ai.
     
    Output
    Output total distance in a line for each testcase.
     
    Sample Input
    2 10 3 2 2 2 8 2 5 1 10 4 1 2 2 8 2 5 1 0 10000
     
    Sample Output
    18 26
     
    Author
    XJZX
     
    Source


    题意:给定一个环,以下标为处为起始点,距离起始点Xi位置种植一颗苹果树,该树有a个苹果,篮子的最大容量为K,那么求摘完全部苹果所需的最短距离。

    当时大致知道方向,但是没想到具体方法。首先贪心左右半边,推出len-k时的最短路程,最后则可以直接走一圈取完。然后再与左右分别取相比较,取较小。

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    #define MAX 100050
    long long sum_left[MAX];
    long long sum_right[MAX];
    int l[MAX],r[MAX];
    
    long long ans;
    int Left,Right;
    int n,k,len,T,a,b;
    
    int main()
    {
        scanf("%d",&T);
        while(T--)
        {
            memset(sum_left,0,sizeof(sum_left));
            memset(sum_right,0,sizeof(sum_right));
    
            scanf("%d%d%d",&len,&n,&k);
            Left=0,Right=0;
            for(int i=0; i<n; i++)
            {
                scanf("%d%d",&a,&b);
                for(int j=0; j<b; j++)
                {
                    if(a*2<len)
                       l[++Left]=a;
                    else
                        r[++Right]=len-a;
                }
            }
            sort(l+1,l+Left+1);
            sort(r+1,r+Right+1);
            for(int i=1; i<=Left; i++)
            {
                if(i<=k)
                    sum_left[i]=l[i];
                else
                    sum_left[i]=sum_left[i-k]+l[i];
            }
            for(int i=1; i<=Right; i++)
            {
                if(i<=k)
                    sum_right[i]=r[i];
                else
                    sum_right[i]=sum_right[i-k]+r[i];
            }
            ans=(sum_left[Left]+sum_right[Right])*2;
    
            for(int i=0; i<=k && i <= Left; i++)
            {
                long long lll = (sum_left[Left-i]+sum_right[max(0,Right-(k-i))])*2;
                ans=min(ans,len+lll);
            }
            printf("%I64d
    ",ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Przz/p/5409828.html
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