Constructing Roads In JGShining‘s Kingdom
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15126 Accepted Submission(s): 4300Problem 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
2
1 2
2 1
3
1 2
2 3
3 1Sample Output
Case 1:
My king, at most 1 road can be built.Case 2:
My king, at most 2 roads can be built.
题意： 上下两条直线上各有ｎ个点，每个点只能连一次， 求没有相交的线段的条数。
由于n较大，需使用ｎ×ｌｏｇｎ的方法求得最长上升子序列。dp[i]表示使得ＬＩＳ长度为ｉ时的最小的结尾。
Accepted Code:
1 /*************************************************************************
2 > File Name: 1025.cpp
3 > Author: Stomach_ache
4 > Mail: [email protected]
5 > Created Time: 2014年07月16日 星期三 23时41分28秒
6 > Propose:
7 ************************************************************************/
8
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17
18 const int MAX_N = 500000+2;
19 int n;
20 int dp[MAX_N], A[MAX_N];
21
22 int
23 main(void) {
24 int c = 1;
25 while (~scanf("%d", &n)) {
26 for (int i = 0; i < n; i++) {
27 int p, r;
28 scanf("%d %d", &p, &r);
29 A[p] = r;
30 }
31 int len = 1;
32 dp[1] = A[1];
33 for (int i = 2; i <= n; i++) {
34 int low = 1, high = len;
35 while (low <= high) {
36 int mid = (low + high) / 2;
37 if (dp[mid] > A[i]) high = mid - 1;
38 else if (dp[mid] < A[i]) low = mid + 1;
39 }
40 dp[low] = A[i];
41 if (low > len) len++;
42 }
43 printf("Case %d:\n", c++);
44 if (1 == len) puts("My king, at most 1 road can be built.");
45 else printf("My king, at most %d roads can be built.\n", len);
46 puts("");
47 }
48
49 return 0;
50 }
Hdu 1025(LIS)