hdu1025 Constructing Roads In JGShining's Kingdom（二分+dp）

【hdu1025 Constructing Roads In JGShining's Kingdom（二分+dp）】的更多相关文章

hdu1025 Constructing Roads In JGShining's Kingdom（二分+dp）

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1025 Problem Description JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines. Half of these cities a…

hdu1025 Constructing Roads In JGShining's Kingdom (nlogn的LIS)

题目链接第一次写nlogn复杂度的LIS,纪念一下. 题目意思是说.有两条平行线.两条平行线都有n个城市,都是从左到右标记为1--n,一条线上是富有城市,一个是贫穷城市.输入n.接下来有n行,p,r表示穷城市p和富有城市r 之间能够建一条路(p的顺序是1--n,一个贫穷城市仅仅相应一个富有城市(弱爆的语文描写叙述能力T_T)),公路不能交叉. 问最多能够建多少条公路. 在别处看到的对nlogn解法的解释吧算是: 时间复杂度:(NlogN): 除了算法一的定义之外,添加一个数组b,b[i]用以表…

HDU 1025 Constructing Roads In JGShining's Kingdom （DP）

Problem Description JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines. Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we ca…

HDU ACM 1025 Constructing Roads In JGShining's Kingdom->二分求解LIS+O(NlogN)

#include<iostream> using namespace std; //BFS+优先队列(打印路径) #define N 500005 int c[N]; int dp[N]; //dp[i]保存的是长度为i的最长不降子序列的最小尾元素 int BS(int n,int x) //二分查找下标,当x比全部元素小时下标为1,比全部元素大时下标为n+1. { int low,high,mid; low=1,high=n; while(low<=high) { mid=(low+h…

hdu-1025 Constructing Roads In JGShining's Kingdom(二分查找)

题目链接: Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21045 Accepted Submission(s): 5950 Problem Description JGShining's kingdom consists of 2n(n is…

DP 60题 -2 HDU1025 Constructing Roads In JGShining's Kingdom

Problem Description JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines. Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we ca…

LIS问题---HDU1025 Constructing Roads In JGShining's Kingdom

发现这个说的比较通俗: 假设存在一个序列d[1..9] = 2 1 5 3 6 4 8 9 7,可以看出来它的LIS长度为5.下面一步一步试着找出它.我们定义一个序列B,然后令 i = 1 to 9 逐个考察这个序列.此外,我们用一个变量Len来记录现在最长算到多少了首先,把d[1]有序地放到B里,令B[1] = 2,就是说当只有1一个数字2的时候,长度为1的LIS的最小末尾是2.这时Len=1然后,把d[2]有序地放到B里,令B[1] = 1,就是说长度为1的LIS的最小末尾是1,d[1]=2…