HDUOJ 1025

原题：

• Problem Description

  …….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.

• 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.
  (空行！！！)
  Case 2:
  My king, at most 2 roads can be built.

解题思路：

• 注意输出格式！！！road和roads！！！
• 说实话这道题的题我读了好久好久，题意理解的(好久×10000)【网上的题意解读杂七杂八很难懂，崩溃ing】，这道题应该这么理解，才能懂应该怎么去解题：
  
  • 输入一个数n表示poor和rich各有n座城市
  • 每一行输入两个数字x、y，分别表示坐标为 x 的poor城市想要坐标为 y 的rich城市的资源
  • poor和rich的所有城市的坐标都是从左到右依次递增的，即两个城市的各个坐标从左到右为1，2，3，……..，n
  • 然后再对应把两个相互需求的p和r城市连接起来，不能有交叉，算出最多能连几条线，即能修多少条路
  • 举个经典的例子：
    
    • 5
    • p：2 3 4 1 5
    • c：2 3 1 4 5
    • 如果在坐标上进行连线，很容易可以看出最多可以连3条线（22，33，55）
    • p：1 2 3 4 5
    •       | |   |
    • c：1 2 3 4 5
    • 1和4，4和1相连都会有交叉
• 所以这个时候应该就有思路了，可以这么去解题：用一个数组a[n]来存每个p城市的坐标和对应的c城市的坐标，a[i] = value : i表示p的坐标，value表示c的坐标，这样p城市的坐标就相当于已经排序好了，现在只需要求出数组a的最长递增子序列，这样肯定能保证求出来的结果不会交叉（p是排好序的，即数组a的下标），而且是最好的结果。

代码：

最长递增子序列优化算法：O（nlogn）

#include <stdio.h>#include <iostream>using namespace std;int dp[500001], temp[500001];int BinarySearch(int tail, int findValue)//二分查找{    int head = 1;    int mid = (head + tail) / 2;    while (mid != 1 && mid != tail)    {        if (dp[mid] == findValue)return mid;        if (dp[mid] > findValue)            tail = mid;        else            head = mid;        mid = (head + tail) / 2;        if (head + 1 == tail && dp[mid] < findValue)            return mid + 1;    }    if (mid == 1 && dp[mid] > findValue)return 1;    else return 2;}int main(){    int temp_p, temp_r, i, dpIndex;    int n, Case = 0;    while (scanf("%d", &n) != EOF)    {        Case++;        for (i = 1; i <= n; i++)        {            scanf("%d%d", &temp_p, &temp_r);            temp[temp_p] = temp_r;        }        dp[1] = temp[1];        dpIndex = 1;        for(i = 2; i <= n; i++)        {            if (temp[i] > dp[dpIndex])                dp[++dpIndex] = temp[i];            else                dp[BinarySearch(dpIndex, temp[i])] = temp[i];        }        if(dpIndex == 1)        {            printf("Case %d:\n", Case);            printf("My king, at most 1 road can be built.\n\n");        }else        {            printf("Case %d:\n", Case);            printf("My king, at most %d roads can be built.\n\n", dpIndex);        }    }}