YJJ's Salesman

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1383    Accepted Submission(s): 483

Problem Description
YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109). YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109), he will only forward to (x+1,y), (x,y+1) or (x+1,y+1).
On the rectangle map from (0,0) to (109,109), there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109), only the one who was from (xk−1,yk−1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.
 
Input
The first line of the input contains an integer T (1≤T≤10),which is the number of test cases.

In each case, the first line of the input contains an integer N (1≤N≤105).The following N lines, the k-th line contains 3 integers, xk,yk,vk (0≤vk≤103), which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.

 
Output
The maximum of dollars YJJ can get.
 
Sample Input
1
3
1 1 1
1 2 2
3 3 1
 
Sample Output
3
 
int a[N],x[N];
int t,n,ret,ans;
void update(int i,int num){
while(i<=n){
a[i]=max(a[i],num);
i+=lowbit(i);
}
}
int query(int n)
{
int ans=;
while(n>)
{
ans=max(ans,a[n]);
n-=lowbit(n);
}
return ans;
}
struct M{
int x,y,ux,uy,w;
}mm[N];
bool cmp3(M a,M b){
if(a.y!=b.y) return a.y<b.y;
return a.x>b.x;
//按列的话要按上式排序,x的排序不能错
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
mem(a,);
gep(i,,n){
scanf("%d%d%d",&mm[i].x,&mm[i].y,&mm[i].w);
x[i]=mm[i].x;
} //都要排序
sort(x+,x++n);//为了lower_bound
sort(mm+,mm++n,cmp3);//为了下面的更新
ret=,ans=;
gep(i,,n){//按列
int k=lower_bound(x+,x++n,mm[i].x)-x;//1~N
ret=query(k-)+mm[i].w;//1~k-1 的最大值+ mm[i].w
update(k,ret);
ans=max(ans,ret);
}
printf("%d\n",ans);
}
return ;
}

HDU 6447的更多相关文章

  1. 2018 CCPC网络赛 1010 hdu 6447 ( 树状数组优化dp)

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=6447 思路:很容易推得dp转移公式:dp[i][j] = max(dp[i][j-1],dp[i-1][j ...

  2. HDU 6447 - YJJ's Salesman - [树状数组优化DP][2018CCPC网络选拔赛第10题]

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6447 Problem DescriptionYJJ is a salesman who has tra ...

  3. HDU 6447 YJJ’s Salesman (树状数组 + DP + 离散)

    题意: 二维平面上N个点,从(0,0)出发到(1e9,1e9),每次只能往右,上,右上三个方向移动, 该N个点只有从它的左下方格点可达,此时可获得收益.求该过程最大收益. 分析:我们很容易就可以想到用 ...

  4. 2018CCPC网络赛

    A - Buy and Resell HDU - 6438 The Power Cube is used as a stash of Exotic Power. There are nn cities ...

  5. BIT+DP

    2018CCPC网络赛 J - YJJ's Salesman HDU - 6447 YJJ is a salesman who has traveled through western country ...

  6. HDOJ 2111. Saving HDU 贪心 结构体排序

    Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

  7. 【HDU 3037】Saving Beans Lucas定理模板

    http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #i ...

  8. hdu 4859 海岸线 Bestcoder Round 1

    http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格 ...

  9. HDU 4569 Special equations(取模)

    Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

随机推荐

  1. siege官方文档(译)(二)

    WHY DO I NEED IT? Siege was written for both web developers and web systems administrators. siege是为了 ...

  2. 为什么会出现lvs+nginx

    一.ngix(应用层 网络七层负载均衡) 1.异步转发,请求数据和相应数据都要经过ngix,ngix和客户端建立连接 2.轮询所有的tomcat服务器,保证请求成功或者最后一台tomcat服务器也请求 ...

  3. 梳理一下我理解的aop

    在看了很多网上的资料和记录之后,我大概捋了下SpringAOP的各种阶段: 基本的advice编程,利用ProxyFactory拿代理类 利用spring把ProxyFactory,advice等be ...

  4. [转]依赖注入框架Autofac的简单使用

    本文转自:http://www.nopchina.net/post/autofac.html 话说nopcommerce底层用到了autofac框架,这里转了一篇文章简单说明一下: Autofac是一 ...

  5. JSP文件上传,好烦啊、、

    <%@ page language="java" import="java.util.*" pageEncoding="UTF-8"% ...

  6. ubuntu下安装无线网卡去驱动Qualcomm-Atheros-QCA9377

    文件(文档和压缩包):http://pan.baidu.com/s/1mhktFT2

  7. 跨平台C++开源代码的两种常用编译方式

    作者:朱金灿 来源:http://blog.csdn.net/clever101 跨平台C++开源代码为适应各种编译器的编译,采用了两种方式方面来适配.一种是makefile方式.以著名的空间数据格式 ...

  8. DRP项目

    DRP(distribution resource planning)分销资源计划是管理企业的分销网络的系统,目的是使企业具有对订单和供货具有快速反应和持续补充库存的能力.解决了随着企业销售规模的逐渐 ...

  9. 当ThreadLocal碰上线程池

    ThreadLocal使用 ThreadLocal可以让线程拥有本地变量,在web环境中,为了方便代码解耦,我们通常用它来保存上下文信息,然后用一个util类提供访问入口,从controller层到s ...

  10. 第14周翻译:SQL Server的阶梯安全级别2

    SQL Server的阶梯安全级别2:身份验证 源自:http://www.sqlservercentral.com/articles/Stairway+Series/109975/ 作者:Don K ...