HDU-1025-Constructing Roads In JGShining's Kingdom

Constructing Roads In JGShining’s Kingdom

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21098 Accepted Submission(s): 5967

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.

看着那么复杂，其实就是求最长上升子序列的长度
数据毕竟多，要用二分查找优化一下。
另一道二分查找动态规划的题，nyoj上的
http://blog.csdn.net/qq_32680617/article/details/50951385

代码

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<stdio.h>
#include<queue>
using namespace std;
//最长上升子序列
//二分dp
int num[500005];//存储数字
int dp[500005];//前i位最长上升子序列长度
int main()
{
    int n;
    int count=1;//计数
    while(~scanf("%d",&n)&&n!=-1)
    {
        int flag1,flag2;//临时接收数据
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&flag1,&flag2);
            num[flag1]=flag2;
        }
        dp[1]=num[1];//初始值
        int length=1;//初始最长上升子序列长度
        for(int i=2; i<=n; i++)
        {
            int left=1;//左边界
            int right=length;//右边界
            while(left<=right)
            {
                int mid=(left+right)/2;
                if(dp[mid]<num[i])
                    left=mid+1;//更新左边界
                else
                    right=mid-1;//更新右边界
            }
            dp[left]=num[i];
            if(left>length)
                length++;//获得最大上升子序列的长度
        }
        printf("Case %d:\n",count++);
        if(length==1)  printf("My king, at most 1 road can be built.\n");
        else    printf("My king, at most %d roads can be built.\n",length);
        printf("\n");
    }
    return 0;
}

因为输出格式的问题，错了四次。。。。