HDU1025(Constructing Roads In JGShining's Kingdom)

这道题我提交了好几次才过的,主要的因为TLE,因为我的算法效率是O(n^2),在网上找到了更高效的是O(n*logn).在这里我附上我两种代码.

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

TLE代码:

Code
#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
    int p;
    int r;
};
node R[500000];

bool comp(node a, node b)
{
    if(a.p != b.p)
        return a.p<b.p;
    else
        return a.r<b.r;
}

int b[500000];
int main()
{
    int n,i,j,max,k=0;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0; i<n; i++)
        {
            b[i]=1;
        }
        for(i=0; i<n; i++)
        {
            scanf("%d %d",&R[i].p,&R[i].r);
        }
        sort(R,R+n,comp);
        max=b[0];
        for(i=1; i<n; i++)
        {
            for(j=0; j<i; j++)
            {
                if(R[j].p<R[i].p && R[j].r<R[i].r && b[i]<b[j]+1)
                {
                    b[i]=b[j]+1;
                }
                if(b[i]>max)
                {
                    max=b[i];
                }
            }
        }
        ++k;
        if(max>1)
            printf("Case %d:\nMy king, at most %d roads can be built.\n\n",k,max);
        else
            printf("Case %d:\nMy king, at most %d road can be built.\n\n",k,max);
    }
    return 0;
}

 

AC代码:

Code
#include<iostream>
using namespace std;
const int Max=500001;
int road[Max],dp[Max];
int main()
{
    int n,i,hight,low,mid,len,k=0;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) 
        { 
            printf("Case %d:\nMy king, at most 0 road can be built.\n\n",++k);
            continue; 
        }
        for(i=0; i<n; i++)
        {
            scanf("%d %d",&hight,&low);
            road[hight]=low;      //有点类似哈希的做法,这里可以省掉排序
        }
        dp[1]=road[1];
        len=1;
        for(i=2; i<=n; i++)
        {
            low=0;
            hight=len;
            while(low<=hight)
            {
                mid=(low+hight)/2;
                if(road[i]>dp[mid])
                    low=mid+1;
                else
                    hight=mid-1;
            }
            dp[low]=road[i];
            if(low>len)
                len++;
        }
        ++k;
        if(len>1)
            printf("Case %d:\nMy king, at most %d roads can be built.\n\n",k,len);
        else
            printf("Case %d:\nMy king, at most %d road can be built.\n\n",k,len);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/chenjunyu/archive/2009/06/05/1497262.html