Constructing Roads In JGShining's Kingdom

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本题目是考察  最长递增子序列的  有n^2     n(logn)  n^2  会超时的

下面两个方法的代码  思路  可以百度LIS  LCS

dp里面存子序列

n(logn)   代码

<span style="font-size:18px;">#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#define N 500000
using namespace std;

int road[N],dp[N],n,len;

int two_part(int *a,int L,int R, int aim)  //二分法找更新的位置
{
    int zz;
    if(len==1&&dp[1]==0) return 1;
    while(L<=R)
    {
        int mid=(L+R)/2;
        if(aim>a[mid]&&aim<a[mid+1]) return mid+1;
        else if(aim>a[mid])   L=mid+1;
        else if(aim<a[mid])  R=mid-1;
    }
    if(aim<=dp[1]) return 1;   //目标数字比第一个数小则更新dp[1]
    return ++len;  //找不到则在末尾更新
}
int ans()
{
    int i;
    for(i=1; i<=n; i++)  //把每个数字在dp数组中更新
    {
        int wz=two_part(dp,1,len,road[i]);
        dp[wz]=road[i];
        // print();
    }
    return len;
}

int main()
{
    int t=1;
    while(scanf("%d",&n)!=EOF)
    {
        int i;
        len=1;
        memset(road,0,sizeof(road));
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
        {
            int ra,rb;
            scanf("%d%d",&ra,&rb);
            road[ra]=rb;
        }
        int answer=ans();
        // printf("%d\n",ans());
        printf("Case %d:\n",t++);
        if(answer==1) printf("My king, at most %d road can be built.\n\n",answer);
        else  printf("My king, at most %d roads can be built.\n\n",answer);
    }
    return 0;
}</span>

<span style="font-size:18px;">
</span>

n^2代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#define N 500000
using namespace std;

int road[N],dp[N],n;

//int two_part(int *a,int L,int R, int aim)
//{
//    while(L<=R)
//    {
//        int mid=(L+R)/2;
//        if(aim==a[mid]) return mid;
//        else if(aim>mid)
//        {
//            L=mid+1;
//        }
//        else
//        {
//            R=mid-1;
//        }
//    }
//    return -1;
//}

int ans()
{
    int sum=0;
    int i,j;
    for(i=0;i<n;i++)
    {
        j=0;
        while(1)
        {
            if(road[i]<dp[j]||!dp[j])
            {
                dp[j]=road[i];
                break;
            }
            j++;
        }
    }
    for(i=0;i<n;i++)
       if(dp[i]) sum++;
    return sum;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int i;
        memset(road,0,sizeof(road));
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
        {
            int ra,rb;
            scanf("%d%d",&ra,&rb);
            road[ra]=rb;
        }
        int answer=ans();
       // printf("%d\n",ans());
        if(answer==1) printf("My king, at most %d road can be built.\n\n",answer);
        else  printf("My king, at most %d roads can be built.\n\n",answer);
    }
    return 0;
}