Constructing Roads In JGShining's Kingdom  HDU1025 题目主要理解要用LCS进行求解! 并且一般的求法会超时!!要用二分!!! 最后蛋疼的是输出格式的注意(直接报错误!!!) #include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; ],dp[]; int main() { ,…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14635    Accepted Submission(s): 4158 Problem Description JGShining's kingdom consists of 2n(n is no mo…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23467    Accepted Submission(s): 6710 Problem Description JGShining's kingdom consists of 2n(n is no mo…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16047    Accepted Submission(s): 4580 Problem Description JGShining's kingdom consists of 2n(n is no mor…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21002    Accepted Submission(s): 5935 Problem Description JGShining's kingdom consists of 2n(n is no mor…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13646    Accepted Submission(s): 3879 Problem Description JGShining's kingdom consists of 2n(n is no mor…

HDOJ(HDU).1025 Constructing Roads In JGShining's Kingdom (DP) 点我挑战题目 题目分析 题目大意就是给出两两配对的poor city和rich city,求解最多能修几条不相交的路.此题可以转化为LIS问题.转化过程如下: 数据中有2列,为方便表述,暂且叫做第一列和第二列. 1.若第一列是是递增的(给出的2个样例都是递增的),那么要想尽可能多的做连线,则那么就需要找出第二列中最长的递增子序列,若出现非递增的序列,那么连线后一定会相交.…

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…

Constructing Roads In JGShining's Kingdom Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27358    Accepted Submission(s): 7782 Problem Description JGShining's kingdom consists of 2n(n is no mor…

http://acm.hdu.edu.cn/showproblem.php?pid=1025 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…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1025 解题报告:先把输入按照r从小到大的顺序排个序,然后就转化成了求p的最长上升子序列问题了,当然按p排序也是一样的.但这题的n的范围是5*10^5次方,所以用n^2算法求 最长上升子序列肯定不行,下面简单介绍一下nlogn时间内求的方法: 从序列里面每次插入一个数,插入到另外一个数组里面,这个数组初始状态是空的,插入一个数时,如果这个数比这个数组里面的任何一个数都大,则直接插入到最后面,否则判断这…

这是最大上升子序列的变形,可并没有LIS那么简单. 需要用到二分查找来优化. 看了别人的代码,给人一种虽不明但觉厉的赶脚 直接复制粘贴了,嘿嘿 原文链接: http://blog.csdn.net/ice_crazy/article/details/7536332 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…

点我看题目 题意 :两条平行线上分别有两种城市的生存,一条线上是贫穷城市,他们每一座城市都刚好只缺乏一种物资,而另一条线上是富有城市,他们每一座城市刚好只富有一种物资,所以要从富有城市出口到贫穷城市,所以要修路,但是不能从富有的修到富有的也不能从贫穷的修到贫穷的,只能从富有的修到贫穷的,但是不允许修交叉路,所以问你最多能修多少条路. 题意 :这个题一开始我瞅了好久都没觉得是DP,后来二师兄给讲了一下才恍然大悟.其实就是用到了DP中的那个最长上升子序列,把题中贫穷城市的坐标当成数组的下标,跟其相连…

主题链接:pid=1025">http://acm.acmcoder.com/showproblem.php?pid=1025 题意:本求最长公共子序列.但数据太多. 转化为求最长不下降子序列.太NB了.复杂度n*log(n). 解法:dp+二分 代码: #include <stdio.h> #include <string.h> #include <vector> #include <string> #include <algorit…

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…

点击打开题目链接 本题目是考察  最长递增子序列的  有n^2     n(logn)  n^2  会超时的 下面两个方法的代码  思路  可以百度LIS  LCS dp里面存子序列 n(logn)   代码 <span style="font-size:18px;">#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #inc…

本题明白题意以后,就可以看出是让求最长上升子序列,但是不知道最长上升子序列的算法,用了很多YY的方法去做,最后还是超时, 因为普通算法时间复杂度为O(n*2),去搜了题解,学习了一下,感觉不错,拿出来分享一下. #include <stdio.h> #include <string.h> #define N 500005 int map[N], dp[N]; int main () { ; while (scanf ("%d", &n) != EOF)…

http://acm.hdu.edu.cn/showproblem.php?pid=1025 题意:富人路与穷人路都分别有从1到n的n个点,现在要在富人点与穷人点之间修路,但是要求路不能交叉,问最多能修多少条. 思路:穷人路是按顺序给的,故求富人路的最长上升子序列即可.由于数据范围太大,应该用O(nlogn)的算法求LIS. #include <stdio.h> #include <algorithm> #include <string.h> using namespa…

最长上升子序列o(nlongn)写法 dp[]=a[]; ; ;i<=n;i++){ if(a[i]>dp[len]) dp[++len]=a[i]; ,dp++len,a[i])=a[i]; } 数组dp[len]描述的是长度为len时的结尾的最小元素. 怎么样才能不交叉呢?i和j相连,i1和j1想连,只有i<i1&&j<j1时才不会交叉,所以让第行单调递增,然后第而行求他的LIS #include<bits/stdc++.h> using names…

题意:贫穷和富有的城市都按顺序1-n排列,需要在中间建造尽可能多的道路,最多可以建造多少条? 解:如果条件这样给出,贫穷的城市按顺序排列,并且按顺序给出每个贫穷城市需要的资源,那么能建造的最多的道路数就是这个顺序中的最长上升子序列的长度. #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include <vector> #include…

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…

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…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1025 题目: 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…

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…

分析: 最长不降子序列,n很大o(n^2)肯定超,想到了小明序列那个题用线段树维护前面的最大值即可 该题也可用二分搜索来做. 注意问题输出时的坑,路复数后加s #include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vecto…

#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…

F. Constructing Roads There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and…

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5227    Accepted Submission(s): 1896 Problem Description There are N villages, which are numbered from 1 to N, and you should b…