HDU1025 最长上升子序列

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Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20548    Accepted Submission(s): 5790


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 call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource.

With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.

Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.

The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones.

But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.

For example, the roads in Figure I are forbidden.



In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^
 

Input
Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.
 

Output
For each test case, output the result in the form of sample.
You should tell JGShining what's the maximal number of road(s) can be built.
 

Sample Input
21 22 131 22 33 1
 

Sample Output
Case 1:My king, at most 1 road can be built.Case 2:My king, at most 2 roads can be built.



这题属于最长上升子序列的问题,解题思路一般的解题思路是两个for循环,不断查找新的最长序列。不过这题时间要求比较高所以用二分查找的方法去进行优化。


开设一个book数组出来,遍历所有的元素,每次的book[i]用来保存长度为i的公共子序列的最小结尾元素。

比如array[20]={2,1,5,3,6,4,8,9,7}

顺序输入数据book[1]=2;然后book[1]=1;

book[2]=5;book[2]=3;book[3]=6;book[3]=4;book[4]=8;book[5]=9;book[4]=7;

这个功能可以用二分查找来实现。


#include <iostream>  #include <cstdio>  #include <cstring>  #include <algorithm>  using namespace std;int  num[50001];int  ans[50001];int main(){int n, k = 1;int len, low, heigh, mid;int tmp1, tmp2;while (scanf("%d", &n) != EOF){for (int i = 0; i<n; i++){scanf("%d %d", &tmp1, &tmp2);num[tmp1] = tmp2;}memset(ans, 0, sizeof(ans));ans[1] = num[1];len = 1;for (int i = 2; i <= n; i++){low = 1;heigh = len;while (low <= heigh){mid = (low + heigh) / 2;if (ans[mid]<num[i])low = mid + 1;elseheigh = mid - 1;}ans[low] = num[i];if (low>len)len++;}printf("Case %d:\n", k);if (len == 1)printf("My king, at most 1 road can be built.");elseprintf("My king, at most %d roads can be built.", len);printf("\n\n");k++;}return 0;}


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