c++之路进阶——斜率优化形如DP[i]=f[j]+x[i](f[j]只与j变量有关)的问题(Print Article)
2016-01-24 17:43
543 查看
参考博文:/article/6002672.html//讲的真的很好,有个小错误,博客里的num全为sum,像我这种弱渣都听懂了。真心点赞!!!
Total Submission(s): 7976 Accepted Submission(s): 2471
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
A single number, meaning the mininum cost to print the article.
[align=left]Sample Input[/align]
5 5
5
9
5
7
5
[align=left]Sample Output[/align]
230
[align=left]Author[/align]
[align=left] [/align]
Xnozero
[align=left]Source[/align]
[align=left]解释: [/align]
[align=left] [/align]
[align=left] 题解:[/align]
1,用一个单调队列来维护解集。
2,假设队列中从头到尾已经有元素a b c。那么当d要入队的时候,我们维护队列的上凸性质,即如果g[d,c]<g[c,b],那么就将c点删除。直到找到g[d,x]>=g[x,y]为止,并将d点加入在 该位置中。
3,求解时候,从队头开始,如果已有元素a b c,当i点要求解时,如果g[b,a]<sum[i],那么说明b点比a点更优,a点可以排除,于是a出队。最后dp[i]=getDp(q[head])。
代码:
顺便吐槽一下,
zhengfeng | We have carefully selected several similar problems for you: 3506 3501 3504 3505 3498
Print Article
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 7976 Accepted Submission(s): 2471
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
A single number, meaning the mininum cost to print the article.
[align=left]Sample Input[/align]
5 5
5
9
5
7
5
[align=left]Sample Output[/align]
230
[align=left]Author[/align]
[align=left] [/align]
Xnozero
[align=left]Source[/align]
[align=left]解释: [/align]
[align=left] [/align]
[align=left] 题解:[/align]
1,用一个单调队列来维护解集。
2,假设队列中从头到尾已经有元素a b c。那么当d要入队的时候,我们维护队列的上凸性质,即如果g[d,c]<g[c,b],那么就将c点删除。直到找到g[d,x]>=g[x,y]为止,并将d点加入在 该位置中。
3,求解时候,从队头开始,如果已有元素a b c,当i点要求解时,如果g[b,a]<sum[i],那么说明b点比a点更优,a点可以排除,于是a出队。最后dp[i]=getDp(q[head])。
代码:
#include<cstdio> #include<cmath> #include<iostream> #define maxn 500100 using namespace std; struct get { int n,m,sum[maxn],a[maxn],dp[maxn],q[maxn]; int getup(int j,int k){return dp[j]+sum[j]*sum[j]-dp[k]-sum[k]*sum[k];}//分子 int getdown(int j,int k){return 2*sum[j]-2*sum[k];}//分母 int getdp(int i,int j) {return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);}//dp[i] get() { while (scanf("%d%d",&n,&m)==2) { for(int i=1;i<=n;i++) scanf("%d",&sum[i]); sum[0]=dp[0]=0; for(int i=1;i<=n;i++)sum[i]+=sum[i-1]; int t=0,w=1; for (int i=1;i<=n;i++) { while (t+1<w&&sum[i]*getdown(q[t+1],q[t])>=getup(q[t+1],q[t])) t++;//维护队列,删除就点之前所有点 dp[i]=getdp(i,q[t]); while (t+1<w&&getup(i,q[w-1])*getdown(q[w-1],q[w-2])<=getup(q[w-1],q[w-2])*getdown(i,q[w-1])) w--;//维护队列,保证队列具有 上凸性质。 q[w++]=i; } printf("%d\n",dp ); } } }get; int main() { get; return 0; }
顺便吐槽一下,
Presentation Error 这种错误你们见过么?我也是醉了!
[align=left]Recommend[/align]zhengfeng | We have carefully selected several similar problems for you: 3506 3501 3504 3505 3498
相关文章推荐
- (模拟) uva 1589 Xiangqi
- 数塔 (HDU_2084) 基础DP
- 最大连续子序列(HDU_1231) 基础DP
- C语言scanf函数详细解释
- memset函数使用详解
- 2014年第五届蓝桥杯B组(C/C++)预赛题目及个人答案(欢迎指正)
- C#调用C++动态库并回调C#函数
- C#调用C++动态库并回调C#函数
- C语言删除目录
- C++字符串相关
- 转载_推荐!国外程序员整理的 C++ 资源大全
- c语言学习之分配不足引起的错误
- 2014年第五届蓝桥杯C/C++程序设计本科B组省赛 地宫取宝(编程大题)
- 2014年第五届蓝桥杯C/C++程序设计本科B组省赛 蚂蚁感冒(编程大题)
- 重学C++ (八) 复制控制
- 【技术研究-泛型1】泛型基础——C++模板
- 打印n对括号的全部n对有效组合
- 2014年第五届蓝桥杯C/C++程序设计本科B组省赛 六角填数(结果填空)
- 2014年第五届蓝桥杯C/C++程序设计本科B组省赛 奇怪的分式(结果填空)
- 身份证号升级