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  • DP(斜率优化):HDU 3507 Print Article

    Print Article

    Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 8199    Accepted Submission(s): 2549

    Problem Description
      Zero has an old printer that doesn't work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
    One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

    M is a const number.
    Now Zero want to know the minimum cost in order to arrange the article perfectly.
     
    Input
      There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.
     
    Output
      A single number, meaning the mininum cost to print the article.
     
    Sample Input
    5 5 5 9 5 7 5
     
    Sample Output
    230
     
      学到了,这样的斜率优化。
      
     1 #include <iostream>
     2 #include <cstdio>
     3 using namespace std;
     4 const int maxn=500010;
     5 long long f[maxn],s[maxn];
     6 int q[maxn]; 
     7 int main()
     8 {
     9     int n,m,front,back;
    10     while(~scanf("%d%d",&n,&m))
    11     {
    12         s[0]=f[0]=0;
    13         for(int i=1;i<=n;i++)scanf("%d",&s[i]);
    14         for(int i=2;i<=n;i++)s[i]+=s[i-1];
    15         
    16         front=1;back=2;
    17         q[front]=0;
    18         
    19         for(int i=1;i<=n;i++)
    20         {
    21             while(front<back-1&&(f[q[front+1]]+s[q[front+1]]*s[q[front+1]])-(f[q[front]]+s[q[front]]*s[q[front]])<=2*s[i]*(s[q[front+1]]-s[q[front]])) front++;
    22             
    23             f[i]=f[q[front]]+(s[i]-s[q[front]])*(s[i]-s[q[front]])+m;
    24             
    25             while(front<back-1&&(s[q[back-1]]-s[q[back-2]])*((f[i]+s[i]*s[i])-(f[q[back-1]]+s[q[back-1]]*s[q[back-1]]))<=(s[i]-s[q[back-1]])*((f[q[back-1]]+s[q[back-1]]*s[q[back-1]])-(f[q[back-2]]+s[q[back-2]]*s[q[back-2]])))back--;
    26             q[back++]=i;
    27         }
    28         printf("%lld
    ",f[n]);
    29     }
    30     return 0;
    31 }
     
    尽最大的努力,做最好的自己!
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  • 原文地址:https://www.cnblogs.com/TenderRun/p/5249644.html
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