zoukankan      html  css  js  c++  java
  • codeforces C. Pearls in a Row map的应用

    C. Pearls in a Row
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    There are n pearls in a row. Let's enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type ai.

    Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.

    Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

    As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

    Input

    The first line contains integer n (1 ≤ n ≤ 3·105) — the number of pearls in a row.

    The second line contains n integers ai (1 ≤ ai ≤ 109) – the type of the i-th pearl.

    Output

    On the first line print integer k — the maximal number of segments in a partition of the row.

    Each of the next k lines should contain two integers lj, rj (1 ≤ lj ≤ rj ≤ n) — the number of the leftmost and the rightmost pearls in the j-th segment.

    Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

    If there are several optimal solutions print any of them. You can print the segments in any order.

    If there are no correct partitions of the row print the number "-1".

    Sample test(s)
    input
    5
    1 2 3 4 1
    output
    1
    1 5
    input
    5
    1 2 3 4 5
    output
    -1
    input
    7
    1 2 1 3 1 2 1
    output
    2
    1 3
    4 7

     题意:将这些算分成一段一段的,每段最多包含两个相同的数字,最多有多少段,和怎么分的段;

     思路:用map记录这个数字在当前这段出现的次数,用队列存当前这段的数字,当有一个数字出现第二次时,清空队列而且保存起点和终点位置;

    AC代码:

    #include <bits/stdc++.h>
    using namespace std;
    const int N=1e5+5;
    int a[5*N];
    struct node
    {
        int a,pos;
    };
    struct po
    {
        int le,ri;
    };
    po ans[5*N];
    int main()
    {
        map<int,int>mp;
        queue<node>qu;
        node x;
       int n;
       int cnt=0;
       scanf("%d",&n);
       for(int i=1;i<=n;i++)
       {
           scanf("%d",&a[i]);
       }
       for(int i=1;i<=n;i++)
       {
           if(mp[a[i]]==0)
           {
               mp[a[i]]++;
               x.a=a[i];
               x.pos=i;
               qu.push(x);
           }
           else
           {
               mp[a[i]]++;
               x.a=a[i];
               x.pos=i;
               qu.push(x);
               int l,r;
               l=qu.front().pos;
               while(!qu.empty())
               {
                   x=qu.front();
                   mp[x.a]--;
                   qu.pop();
               }
               r=x.pos;
               cnt++;
               ans[cnt].le=l;
               ans[cnt].ri=r;
           }
       }
       if(!cnt)cout<<"-1"<<endl;
       else
       {
           printf("%d
    ",cnt);
           for(int i=1;i<cnt;i++)
           {
               printf("%d %d
    ",ans[i].le,ans[i].ri);
           }
           printf("%d %d
    ",ans[cnt].le,n);
       }
        return 0;
    }
  • 相关阅读:
    我的第一个可用的Windows驱动完成了
    据说是一种很古老的方法
    起一卦,测今天工作,问题不少
    起一卦,找房子,马上没房子住了
    哈哈哈哈,我竟然发现了个MSDN里面的笔误
    起一卦,看现在我的工程进度怎么样。
    起卦帮同学看工作,应了。
    2012年10月17日帮朋友算得第一卦
    2013年1月13日帮朋友测的第二卦,有些地方没看出来
    bzoj2588 Spoj 10628. Count on a tree
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5158545.html
Copyright © 2011-2022 走看看