hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
来源:互联网 发布:小猫软件scratch下载 编辑:程序博客网 时间:2024/05/16 19:23
题意:
由n'个数组成的环形序列a1,a2...an,其中a1左边是an,an右边是a1,,求该序列中长度不超过k的子序列的最大和。
思路:
将a1..an复制到a1+n...a2*n,然后用sum[i]记录前i项和,那么sum[i]-sum[i-m]就是长度为m的子序列的和,用单调队列维护下标,具体过程如下。
代码:
/* * main.cpp * * Created on: 2015年10月21日 * Author: chen */#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<memory.h>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff //INT_MAX#define inf 0x3f3f3f3f //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++) //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define LL __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }using namespace std;template<class T>T Mint(T a, T b, T c) {if (a>b) {if (c>b)return b;return c;}if (c>a)return a;return c;}template<class T>T Maxt(T a, T b, T c) {if (a>b) {if (c>a)return c;return a;}else if (c > b)return c;return b;}const int maxn=200010;int T,n,m;int a[maxn];int sum[maxn];int main() {#ifndef ONLINE_JUDGE freopen("test.in","r",stdin); freopen("test.out","w",stdout);#endif sf(T); while(T--){ sfd(n,m); for2(i,1,n){ sf(a[i]); a[i+n]=a[i]; } sum[0]=0; for(int i=1;i<n+m;i++){//预处理前i项和 sum[i]=sum[i-1]+a[i]; } int ans=-inf,s,e; deque<int>q; for2(i,1,n+m-1){ while(!q.empty()&&sum[i-1]<sum[q.back()])//维护单调性,如果第i项比第i-1项和小,则第i-1项不是最优,删除。 q.pop_back();//比如sum[i]<sum[i-1],则A-sum[i]>A-sum[i-1]. while(!q.empty()&&q.front()<i-m)//如果子序列长度大于k,则删除首项。 q.pop_front(); q.push_back(i-1);//插入队尾 int t=sum[i]-sum[q.front()]; if(t>ans){//更新最大值和下标 ans=t; s=q.front()+1; e=i; } } printf("%d %d %d\n",ans,s,e>n?e-n:e); }return 0;}
0 0
- HDU Max Sum of Max-K-sub-sequence(单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence (单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- [ACM] hdu 3415 Max Sum of Max-K-sub-sequence (单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence【单调队列】
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence[单调队列优化dp]
- 【单调队列】hdu 3415 Max Sum of Max-K-sub-sequence
- hdu 3415 Max Sum of Max-K-sub-sequence 单调队列dp
- HDU 3415(Max Sum of Max-K-sub-sequence-单调队列优化DP)
- HDU--杭电--3415--Max Sum of Max-K-sub-sequence--暴力或单调队列
- 一次memcached的排查
- poj 1986 Distance Queries(LCA离线Tarjan算法)
- TextView和EditText小结
- 没有躲过的坑--This function or variable may be unsafe.
- 图(邻接矩阵)的深度、广度优先遍历
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- [kuangbin带你飞]专题一 简单搜索 I - Fire Game FZU 2150
- 黑马程序员--自学笔记--集合(其一)
- nyoj 摆方格 1087 (数学规律)
- 后厨-物流
- 农夫过河问题(图的邻接矩阵)
- hdu 3951(博弈,成环)
- 使用MP4V2开源库将AAC打包到MP4文件中
- Android 网络编程