标签：style blog color os io for ar cti
to get the ans of how many roads at most that can be built between two line without intersection of roads,we need sort the input sequence at ont edge and then deal with anther line with LIS
way,meanwhile, not forget using the nlogn way.
1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 const int N=500009; 5 struct LINE 6 { 7 int from,to; 8 friend bool operator<(LINE A,LINE B) 9 { 10 return A.from<B.from; 11 } 12 }; 13 LINE node[N]; 14 int y[N]; 15 int c[N]; 16 int MID(int a[],int value,int size) 17 { 18 int l=1,r=size; 19 int mid=(l+r)/2; 20 while(l<r)//1 3 5 7 9 3 21 { 22 if(a[mid]<value&&value<=a[mid+1])return mid+1; 23 else if(value<=a[mid])r=mid; 24 else l=mid+1; 25 mid=(l+r)/2; 26 } 27 } 28 int LIS(int len) 29 { 30 int size=1; 31 int zhong=0; 32 c[1]=y[1]; 33 for(int i=2;i<=len;i++)//7 6 4 9 11 14 17 3 15 20 4 5 6 7 8 34 {//3 9 11 15 17 20 35 if(y[i]<=c[1])c[1]=y[i]; 36 else if(y[i]>c[size]) 37 c[++size]=y[i]; 38 else 39 { 40 zhong=MID(c,y[i],size); 41 c[zhong]=y[i]; 42 } 43 } 44 return size; 45 } 46 int main() 47 { 48 int n; 49 int cnt=1; 50 while(cin>>n) 51 { 52 for(int i=0;i<n;i++) 53 scanf("%d %d",&node[i].from,&node[i].to); 54 sort(node,node+n);//sort one side by ascending direction 55 for(int i=0;i<n;i++) 56 y[i+1]=node[i].to; 57 int ans=LIS(n); 58 cout<<"Case "<<cnt<<‘:‘<<endl; 59 cout<<"My king, at most "<<ans; 60 if(ans>1)cout<<" roads can be built."<<endl; 61 else cout<<" road can be built."<<endl; 62 cnt++; 63 cout<<endl; 64 } 65 return 0; 66 }
It has used the longest increasing subsequence algorithm way,and to get it solved after you have learned it.
input : with a n,represent n pairs of integers followed.And n pairs of intergers followed
output:
Cast i:
My king, at most n road(s) can be built
A sample below:
11
2 5
3 6
4 2
5 1
6 9
7 11
8 7
9 12
10 8
11 4
12 10
Case 1:
My king, at most 6 road can be built.
标签：style blog color os io for ar cti
原文地址：http://www.cnblogs.com/sytu/p/3916549.html
