hdu 1025 Constructing Roads In JGShining’s Kingdom 【dp+二分】

题目链接：http://acm.acmcoder.com/showproblem.php?pid=1025

题意：本求最长公共子序列，但数据太多。转化为求最长不下降子序列。太NB了。复杂度n*log(n).

解法：dp+二分

代码：

#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <fstream>
#include <set>
#include <map>
#include <math.h>

using namespace std;

const int MAXN = 500010;

int n, pos;
int a[MAXN];
int dp[MAXN];
int h, k;

int search(int num,int low,int high)
{
    int mid;
    while (low <= high)
    {
        mid = (low + high) / 2;
        if (num >= dp[mid]) low = mid + 1;
        else high = mid - 1;
    }
    return low;
}

int main()
{
    int cases = 1;
    while (~scanf("%d", &n))
    {
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d", &h, &k);
            a[h] = k;
        }
        memset(dp, 0, sizeof(dp));
        dp[0] = -1; dp[1] = a[1];
        int len = 1;
        //  n*log(n) 求解

        for (int i = 2; i <= n; i++)
        {
            if (a[i] >= dp[len])
            {
                len = len + 1;
                dp[len] = a[i];
            }
            else
            {
                pos = search(a[i],1,len);
                dp[pos] = a[i];
            }
        }
        printf("Case %d:\n",cases++);
        if (len == 1)
            printf("My king, at most %d road can be built.\n\n",len);
        else
            printf("My king, at most %d roads can be built.\n\n",len);
    }
    return 0;
}