HDU 1025 Constructing Roads In JGShining's Kingdom

 

http://acm.hdu.edu.cn/showproblem.php?pid=1025

题目有一定迷惑性，实际就是求LIS，我原来掌握的朴实的O(n^2)算法果断超时，新学了一种二分dp O(nlog(n))的算法，直接上模板了，还要多多体会啊

View Code

#include <iostream>

using namespace std ;

int dp[10001],p[50001];

int LIS(int n)

{

    int l,r,m,i,tail = 0;

    for ( dp[ ++ tail ] = p[ 1 ],i = 2 ; i <= n ; ++ i ) 

    {

        if ( dp[ tail ] <= p[ i ] ) 

        {

            dp[ ++ tail ] = p[ i ];

            continue;

        }

        for ( m=((r=tail)+(l=1)>>1) ; l < r ; m=(l+r)>>1 )

            if ( dp[ m ] <= p[ i ] ) l = m+1;

            else r = m;

        dp[ m ] = p[ i ];

    }

    return tail;

}

inline bool scan_d(int &num) 

{

        char in;bool IsN=false;

        in=getchar();

        if(in==EOF) return false;

        while(in!='-'&&(in<'0'||in>'9')) in=getchar();

        if(in=='-'){ IsN=true;num=0;}

        else num=in-'0';

        while(in=getchar(),in>='0'&&in<='9'){

                num*=10,num+=in-'0';

        }

        if(IsN) num=-num;

        return true;

}

int main()

{

    int n,nCase=1;

    while(scan_d(n))

    {

        for(int i=0;i<n;i++)

        {

            int x,y;

            scan_d(x);

            scan_d(y); 

            p[x]=y;

        }

        int ans=LIS(n);

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

        if(ans==1)

            puts("My king, at most 1 road can be built.");

        else

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

        putchar('\n');

    }

    return 0;

}