經典題。。。最長非遞減序列的n*log(n)求法。。。orz...

 1 #include<iostream>
 2 const int N=500007;
 3 using namespace std;
 4 int city[N];
 5 int dp[N];//dp[i]儲存的是長度為i的最長不降子序列的最小尾元素
 6 
 7 //二分查詢返回num在dp中的位置
 8 int Search(int dp[],int len,int num){
 9     int low=1,high=len;
10     while(low<=high){
11         int
 
 mid=(low+high)/2;
12         if(num==dp[mid])return mid;
13         else if(num<dp[mid])high=mid-1;
14         else if(num>dp[mid])low=mid+1;
15     }
16     return low;
17 }
18 
19 int main(){
20     int n,_case=1;
21     while(~scanf("%d",&n)){
22         int a,b;
23         for(int i=1;i<=n;i++){
 
24             scanf("%d%d",&a,&b);
25             city[a]=b;
26         }
27         dp[0]=-1,dp[1]=city[1];
28         int len=1;
29         //n*log(n)求最長單調非遞減序列
30         for(int i=1;i<=n;i++){
31             int j=Search(dp,len,city[i]);
32             dp[j]=city[i];//每次都要更新dp
33             if
 
(j>len)len++;
34         }
35         printf("Case %d:\n",_case++);
36         if(len==1){
37             printf("My king, at most %d road can be built.\n\n",len);
38         }else 
39             printf("My king, at most %d roads can be built.\n\n",len);
40     }
41     return 0;
42 }
43 
44 
45