hdu 1025 Constructing Roads In JGShining's Kingdom dp

听说这是个LIS的题，画了画图，发现是这样的..

#include <stdio.h>

int ar[500005],vis[500000];

int main()
{
    int n,i,j,k,m,p,r,l=1;
    while (scanf("%d",&n)!=EOF)
    {
        for (i=0;i<n;i++)
        {
            scanf("%d%d",&p,&r);
            ar[p]=r;
        }
        for (i=1,j=0,m=0;i<=n;i++)
        {
            if(ar[i]>m)
            {
                m=ar[i];
                vis[++j]=m;
                continue;
            }
            for (k=j-1;;k--)
            {
                if(ar[i]>vis[k]||k==0)
                {
                    vis[k+1]=ar[i];
                    m=vis[j];
                    break;
                }
            }
        }
        printf("Case %d:\n",l++);
        printf("My king, at most %d road%scan be built.\n\n",j,j==1?"":"s ");
    }
    return 0;
}

看到个二分解LIS的代码，学习了.

#include<iostream>

using namespace std;

int ar[500001],dp[500001];

int main()
{
    int i,n,c=1,a,b,len,low,up,mid;
    while(scanf("%d",&n)!=-1)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a,&b);
            ar[a]=b;//表示a城市通向b城市
        }
        dp[1] = ar[1];//初始化dp[1]的值
        len = 1;
        for(i=1;i<=n;i++)
        {
            //二分查找
            low = 1;
            up = len;
            while(low<=up)
            {
                mid = (low+up)/2;
                if(dp[mid]>=ar[i])up = mid -1;
                else low = mid+1;
            }
            dp[low] = ar[i];
            if(low > len)len++;
        }
        printf("Case %d:\n",c++);
        printf("My king, at most %d road%scan be built.\n\n",len,len==1?"":"s ");
    }
    return 0;
}