hdu--（1025）Constructing Roads In JGShining's Kingdom（dp/LIS+二分）

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16047    Accepted Submission(s): 4580

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.

 

 

Author

JGShining（极光炫影）

 

这道题，意思很明了，每一个点对应一个点，不存在一个点对多个点的情况，也不存在多个点对一个点的情况，且不能存在相交线。

问你连那些点使得线条最多。

又给出的a b,我们不难构造出str[a]=b,a为1 -n 给的升序，所以我们只需要对于str[a]这个数组内的值进行单调递增子序列处理就可以了....简单明了

代码：

 1 //#define LOCAL

 2 #include<cstdio>

 3 #include<cstring>

 4 const int maxn=50005;

 5 const int inf=0x3f3f3f3f;

 6 int str[maxn],dp[maxn],sac[maxn];

 7 int res;

 8 

 9 int binary(int v,int n)

10 {

11     int ll=1,rr=n,mid;

12     while(ll<=rr)

13     {

14       mid=ll+(rr-ll)/2;

15       if(sac[mid]<=v&&sac[mid]!=-1)

16         ll=mid+1;

17       else

18         rr=mid-1;

19     }

20     return ll;

21 }

22 void LIS(int n)

23 {

24       res=1;

25 //    for(int i=1;i<=n;i++)

26 //      sac[i]=inf;

27     memset(sac,-1,sizeof(int)*(n+1));

28    for(int i=1;i<=n;i++)

29    {

30        dp[i]=binary(str[i],res);

31        if(res<dp[i])

32            res=dp[i];

33        if(str[i]<sac[dp[i]]||sac[dp[i]]==-1)

34            sac[dp[i]]=str[i];

35    }

36 }

37 int main()

38 {

39   #ifdef LOCAL

40   freopen("test.in","r",stdin);

41   #endif

42 

43   int n,i,p,r,ct=1;

44   while(scanf("%d",&n)!=EOF)

45   {

46       for( i=1;i<=n;i++){

47       scanf("%d%d",&p,&r);

48         str[p]=r;

49       }

50         LIS(n);

51       printf("Case %d:\n",ct++);

52       if(res==1)

53        printf("My king, at most 1 road can be built.\n");

54     else

55        printf("My king, at most %d roads can be built.\n",res);

56     printf("\n");

57   }

58   return 0;

59 }

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