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

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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 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 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.

注意区分road和roads

思路：

/*如果存在一个序列d[1..9] = 2 1 5 3 6 4 8 9 7。能够看出来它的LIS长度为5。

然后应该发现一件事情了：在B中插入数据是有序的，并且是进行替换而不须要挪动——也就是说，我们能够使用二分查找。将每个数字的插入时间优化到O(logN)~~~~~于是算法的时间复杂度就减少到了O(NlogN)～

*/

代码例如以下：

#include <cstdio>
#include <cstring>
const int maxn = 500047;
int dp[maxn], ans[maxn], n;
int LIS()
{
    dp[1] = ans[1]; //dp[i]用来存储长度为i时的最小值
    int len = 1;
    for (int i = 2; i <= n; i++)
	{
        int left = 1, right = len;
        while (left <= right)//二分查找应该更新的元素
		{
            int m = (left + right)/2;
            if (ans[i] > dp[m])
				left = m + 1;
            else
				right = m - 1;
        }
        dp[left] = ans[i];
        if (left > len)
			len = left;
    }
    return len;
}
int main()
{
    int x, y,k = 1;
    while (~scanf("%d", &n))
	{
        for (int i = 0; i < n; i++)
		{

            scanf("%d%d", &x, &y);
            ans[x] = y;
        }
        memset(dp, 0, sizeof(dp));
        int len = LIS();
        printf("Case %d:\n",k);
        if(len==1)
			printf("My king, at most 1 road can be built.");
        else
			printf("My king, at most %d roads can be built.",len);
        printf("\n\n");
        k++;
    }
    return 0;
}

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