STL or Force --- CSU 1553: Good subsequence

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Good subsequence

Problem's Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1553

Mean:

给你一个长度为n的序列和一个值k，让你找出一个子序列，满足在这个子序列中max-min的值<=k，求这个子序列最长的长度。

analyse:

这题做法很多，直接暴力枚举每一个数为起点。

Time complexity: O(n)

Source code:

方法一（暴力）：

//  Memory   Time//  1347K     0MS//   by : crazyacking//   2015-03-30-16.02#include<map>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<cstdlib>#include<cstring>#include<climits>#include<iostream>#include<algorithm>#define MAXN 1000010#define LL long longusing namespace std;int n,k;vector<int> ve;int main(){        ios_base::sync_with_stdio(false);        cin.tie(0);//      freopen("C:\\Users\\Devin\\Desktop\\cin.cpp","r",stdin);//      freopen("C:\\Users\\Devin\\Desktop\\cout.cpp","w",stdout);        while(cin>>n>>k)        {                ve.clear();                for(int i=0;i<n;++i)                {                        int tmp;                        cin>>tmp;                        ve.push_back(tmp);                }                int ans=1;                for(int i=0;i<n;++i)                {                        int cnt=1;                        int maxx=ve[i];                        int minn=ve[i];                        for(int j=i+1;j<n;++j)                        {                                if(ve[j]>=maxx)                                {                                        maxx=ve[j];                                }                                if(ve[j]<=minn)                                {                                        minn=ve[j];                                }                                if(maxx-minn>k) break;                                cnt++;                        }                        if(cnt>ans) ans=cnt;                }                cout<<ans<<endl;        }        return 0;}/**/

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方法二(STL):

做法很巧妙，用一个multiset来维护：加入当前这个数后满足条件的连续子序列，也就是说每一轮循环set中的元素都是满足条件的。

//  Memory   Time//  1347K     0MS//   by : crazyacking//   2015-03-30-15.53#include<map>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<set>#include<string>#include<cstdlib>#include<cstring>#include<climits>#include<iostream>#include<algorithm>#define MAXN 1000010#define LL long longusing namespace std;int n,k;multiset<int> se;vector<int> ve;int main(){        ios_base::sync_with_stdio(false);        cin.tie(0);//      freopen("C:\\Users\\Devin\\Desktop\\cin.cpp","r",stdin);//      freopen("C:\\Users\\Devin\\Desktop\\cout.cpp","w",stdout);        while(cin>>n>>k)        {                se.clear();                ve.clear();                for(int i=0;i<n;++i)                {                        int tmp;                        cin>>tmp;                        ve.push_back(tmp);                }                int ans=1;                for(int i=0,j=0;i<n;++i)                {                        se.insert(ve[i]);                        for(;*se.rbegin()-*se.begin()>k;j++)                        {                                se.erase(ve[j]);                        }                        if(i-j+1>ans) ans=i-j+1;                }                cout<<ans<<endl;        }        return 0;}/**/

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