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;}