HDU 1025 - Constructing Roads In JGShining's Kingdom(最长上升子序列)

Constructing Roads In JGShining’s Kingdom

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23439 Accepted Submission(s): 6705

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.

Hint

Huge input, scanf is recommended.

题意:
求最长上升子序列长度.

解题思路:
分出两个数组,A数组存原数组,B数组存最长上升子序列.
实现过程是从头遍历A数组,如果遍历到的这个数大于等于B数组的最后一个,那么就将该数添加到B数组的末尾,否则在B数组中找到第一个大于该数的数,并将其替换(查找过程用二分实现,因为B数组在操作过程中一直保持有序).

AC代码:

#include<stdio.h>
const int maxn = 500010;
int a[maxn];
int b[maxn];
int search(int num,int low,int high)
{
    while(low<=high)
    {
        int mid = (low+high)>>1;
        //if(num == b[mid]) return mid; /*如果是严格上升*/
        if(num >= b[mid])   low = mid+1;
        else                high = mid-1;
    }
    return low;
}
int dp(int n)
{
    b[1] = a[1];
    int len = 1;
    for(int i = 2;i <= n;i++)
    {
        if(a[i] >= b[len])
            b[++len] = a[i];
        else
            b[search(a[i],1,len)] = a[i];
    }
    return len;
}
int main()
{
    int flag = 1;
    int n;
    while(~scanf("%d",&n))
    {
        int index;
        int num;
        for(int i = 1;i <= n;i++)
        {
            scanf("%d%d",&index,&num);
            a[index] = num;
        }
        int res = dp(n);
        printf("Case %d:\n",flag++);
        if(res == 1)  printf("My king, at most 1 road can be built.\n\n");
        else            printf("My king, at most %d roads can be built.\n\n",res);
    }
    return 0;
}