zoukankan      html  css  js  c++  java
  • HDU 3507 Print Article

    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
     
    题解:
    斜率优化dp板子题.单调队列维护 x满足递增 非常好写
    一个结论:满足条件的点一定在一个凸包的底端,所以我们维护凸包底端的点即可
     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstdlib>
     4 #include <cstring>
     5 #include <cstdio>
     6 #include <cmath>
     7 using namespace std;
     8 typedef long long ll;
     9 const int N=500005;
    10 int gi(){
    11     int str=0;char ch=getchar();
    12     while(ch>'9' || ch<'0')ch=getchar();
    13     while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,ch=getchar();
    14     return str;
    15 }
    16 int n,m,a[N],q[N];ll sum[N],f[N];
    17 ll fy(int i,int j){
    18     return sum[i]*sum[i]+f[i]-sum[j]*sum[j]-f[j];
    19 }
    20 ll fx(int i,int j){
    21     return ((sum[i]-sum[j])<<1);
    22 } 
    23 void work(){
    24     int l=1,r=1,j,k;
    25     q[1]=0;
    26     for(int i=1;i<=n;i++)a[i]=gi(),sum[i]=sum[i-1]+a[i];
    27     for(int i=1;i<=n;i++){
    28         while(l<=r-1){
    29             j=q[l];k=q[l+1];
    30             if(fy(j,k)>=sum[i]*fx(j,k))l++;
    31             else break;
    32         }
    33         f[i]=f[q[l]]+m+(sum[i]-sum[q[l]])*(sum[i]-sum[q[l]]);
    34         while(l<=r-1){
    35             j=q[r];k=q[r-1];
    36             if(fy(i,j)*fx(j,k)<=fy(j,k)*fx(i,j))r--;
    37             else break;
    38         }
    39         q[++r]=i;
    40     }
    41     printf("%lld
    ",f[n]);
    42 }
    43 int main()
    44 {
    45     while(~scanf("%d%d",&n,&m))
    46     work();
    47     return 0;
    48 }
  • 相关阅读:
    MySQL中 Data truncated for column 'xxx'解决方法
    JAVA中循环删除list中元素的方法总结
    Java 键盘输入数字(空格隔开) 将数字存入数组
    linux查看服务器并发连接数
    解决 httpclient 下 Address already in use: connect 的错误
    知识点--实际开发中软引用或弱引用的使用场景
    无序hashset与hashmap让其有序
    bool的值分别为0,1;那哪个代表true哪个代表false?
    jquery-ui autocomplete在模态框(model)中,出不来
    vue-Treeselect实现组织机构(员工)下拉树的功能
  • 原文地址:https://www.cnblogs.com/Yuzao/p/7237135.html
Copyright © 2011-2022 走看看