HDU1025 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): 11914    Accepted Submission(s): 3393

Problem DescriptionJGShining'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. ^_^ 

InputEach 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. 

OutputFor 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

21 22 131 22 33 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.HintHuge input, scanf is recommended. 

 

AuthorJGShining（極光炫影）

解題思路：一看就知道是求最長子序列有木有，動歸有木有。好吧，我承認我逾時過，題目資料量比較大an integer n(1 ≤ n ≤ 500,000).如果用經典的最長子序列求法，會逾時。
            對於題意，處理時可以化簡，每對城市（x即缺資源的，y即自願豐富的）輸入後可以直接處理到數值ct[]中，ct[x]=y.這樣就把問題轉變成求序列ct[]的最長遞增子序列。
           用LIS演算法求最長子序列，不用曆遍前面各子序列的最長子序列，節約時間。直接用棧儲存當前最長子序列的值，最後最長子序列的長度為棧的長度。這裡我用數組類比棧。對於每個元素Y（x已經遞增），檢索它與數組中已儲存的最長子序列的元素的關係（數組最後元素road[max1-1]）：（1）若ct[x]>road[max1-1],則直接將ct[x]儲存到數組後面即可(入“棧”)，這裡不用解釋啦，可以直接增長最長子序列的長度。（2）若（1）不成立，則尋找road中第一個（從“頭”開始）比ct[x]大的元素（二分法），並替換它即可（置換），可增加後面元素的“可入棧性”。這個不好直接講清楚，我們上案例：
ct[x ]
x    1    2    3    4    5    6    7    8
y    2    5    1    3    4    6    8    7

則對於棧的類比數組road[ max1 ]：  

最後，最長子序列為max1-1（每存入一個元素都max++）。                           

#include<stdio.h>int ct[500005];int road[500005];int lown(int i,int x)   //二分尋找第一個比x大的元素在road中的下標位置{    int j=0;    while(j<=i)    {        int m=(i+j)>>1;        if(x<road[m])            i=m-1;        else if(x>road[m])            j=m+1;        else            return m;    }    return j;}int main(){    int t=1;    int n;    int i,j;    int max1;    while(scanf("%d",&n)!=EOF)    {        max1=0;        int x,y;        for(i=0;i<n;i++)        scanf("%d%d",&x,&y),ct[x]=y;   //轉化為求ct的最長遞增子序列        road[max1++]=ct[1];        for(i=2;i<=n;i++)        {            if(ct[i]>road[max1-1])    //入“棧”                road[max1++]=ct[i];            else                road[lown(max1-1,ct[i])]=ct[i];     //置換        }        printf("Case %d:\n",t++);        printf("My king, at most %d road",max1);        if(max1!=1)        printf("s");    //最長子序列長度大於1時，要用複數形式（加‘s’）        printf(" can be built.\n\n");    }    return 0;}