Jia Xin Yi sister and twin Cheese bun to simulate the QDUOJ

OJ point I entered the details of the subject

The meaning of problems

Give you a bunch of number, so that you find the length of the longest substring, the string figures which did not meet recurring.

Problem-solving ideas

Using an array of tags to mark the position of the first occurrence of each number, then the following simulation, code implementation details see it.

Han Han was himself, did not think the need for discrete, for the first time after the submission aware of the need to return immediately into the heart of discrete After Runtime Error, but no time, hey, too bad. Of course, when solving the problem, their own ideas is not very clear, leading to Debug a long time.

Code

//这个是自己思路清晰后写的代码，思路将相当与两个指针
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector> 
using namespace std;
const int maxn=1e6+7;
vector<int> v;
int a[maxn], vis[maxn];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        while(!v.empty()) v.pop_back() ; //清空上次的数据
        for(int i=1; i<=n; i++)
            vis[i]=0; //初始化
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
            v.push_back(a[i]);  
        }       
        sort(v.begin() , v.end() );
        v.erase(unique(v.begin() , v.end() ), v.end() ); //这行和上一行是进行离散化
        for(int i=1; i<=n; i++)
        {
            a[i]=lower_bound(v.begin() , v.end() , a[i])-v.begin()+1;
            if(vis[a[i]]==0)//这里只记录第一次数字出现的位置 
                vis[a[i]]=i;
        }
        int i, begin=1, ans=0;//begin相当于左指针
        for(i=1; i<=n; i++)//这里的i相当于右指针
        {
            if(vis[a[i]]!=i)
            {
                if(vis[a[i]] < begin) //如果第一次出现的位置，小于左指针指向的位置，那么就可以直接更改标记数组就行了，这里可以用笔和纸模拟一下就可以了。
                {
                    vis[a[i]]=i;
                }
                else 
                {
                    ans=max(ans, i-begin); //i-begin是长度
                    begin=vis[a[i]]+1;//更改左边的位置
                    vis[a[i]]=i; //更改标记数组
                }
            }
        }
        ans=max(ans, i-begin); //这里最后一次也需要进行判断一次
        printf("%d\n", ans); 
    } 
    return 0;
}

//这里加了离散化就过来，还是自己太菜了！
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
int a[maxn];
int vis[maxn];
vector<int> v;
int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        while(!v.empty()) v.pop_back() ;
        for(int i=1; i<maxn; i++)
            vis[i]=0; 
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
            v.push_back(a[i]);
        }
        sort(v.begin(), v.end());
        v.erase( unique(v.begin(), v.end() ), v.end() );
        for(int i=1; i<=n; i++)
        {
            int tmp=lower_bound(v.begin(), v.end() , a[i]) - v.begin() +1;
            a[i]=tmp;
            if(vis[tmp]==0)
                vis[tmp]=i;
        }
        int ans=0, tmp=0, begin=1, flag=0;
        for(int i=1; i<=n; i++)
        {
            flag=0;
            if(vis[a[i]]!=i && vis[a[i]]!=0)
            {
                if(vis[a[i]]>=begin)
                {
                    ans=max(tmp, ans);
                    tmp++;
                    if(vis[a[i]]==begin)
                        tmp=tmp-1;
                    else 
                        tmp=tmp-vis[a[i]]+begin-1;
                    begin=vis[a[i]]+1;
                    flag=1;
                }
                vis[a[i]]=i;
                if(flag==0)
                    tmp++;
            }
            else{
                tmp++;
            }
        }
        ans=max(ans, tmp);
        printf("%d\n", ans);
    }
    return 0;
}