CF620C Pearls in a Row
题目描述
There are nn pearls in a row. Let's enumerate them with integers from 11 to nn from the left to the right. The pearl number ii has the type a_{i}a**i .
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/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
输入格式
The first line contains integer nn ( 1<=n<=3·10^{5}1<=n<=3⋅105 ) — the number of pearls in a row.
The second line contains nn integers a_{i}a**i ( 1<=a_{i}<=10^{9}1<=a**i<=109 ) – the type of the ii -th pearl.
输出格式
On the first line print integer kk — the maximal number of segments in a partition of the row.
Each of the next kk lines should contain two integers l_{j},r_{j}l**j,r**j ( 1<=l_{j}<=r_{j}<=n1<=l**j<=r**j<=n ) — the number of the leftmost and the rightmost pearls in the jj -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".
题意翻译
现在有N个数，你的任务是将这N个数尽可能切割成多段。每一段必须包括两个相同的数。
数据输入量比较大，建议使用scanf和printf。
Input 单组测试数据。数据第一行为N(1 ≤ N ≤ 3·105) 。
数据的第二行包括N个数ai(1 ≤ ai ≤ 109) 。
Output 输出的第一行为尽可能切割的最大段数K。
接下来K行，每行为两个整数lj, rj (1 ≤ lj ≤ rj ≤ n) ，表示切割的区间范围
如果存在多个合法的切割方法，输出任意一个即可。
如果不能切割成合法的情况，输出"-1".
输入输出样例
输入 #1复制
输出 #1复制
输入 #2复制
输出 #2复制
输入 #3复制
输出 #3复制
题解：
2019.11.11光棍节模拟赛T1 100分场可海星
这是我自从参加(JDFZ\,\,CSP-S)模拟赛之后第一次秒切T1的场！！炒鸡鸡冻！
我觉得思路真的特别好想：
一边输入一边标记出现的数，如果碰到已经打上标记的数就清空标记数组，(++cnt)并且开一个(last)变量存区间分割点。
但是(bool)数组开(1e9)就会炸...
咋办呢？
我们发现(STL)给我们提供了一个特别好用的容器(set)，专门用来维护元素重复的集合。
里面提供了一个特别好用的函数(find())，能返回等于元素的迭代器。如果返回的迭代器不等于(s.end())，说明有这个元素，就可以加进答案里。
在(CF)交的时候还出现了一种始料未及的情况，就是这种做法可能会出现不覆盖满的情况，所以退出来的时候要暴力地把(ans[cnt].r)变成(n)。
完整代码如下：
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
const int maxn=3*1e5+10;
int n,last,cnt;
int a[maxn];
struct node
{
int l,r;
}ans[maxn];
set<int> s;
set<int>::iterator it;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
last=1;
for(int i=1;i<=n;i++)
{
if((it=s.find(a[i]))!=s.end())
{
ans[++cnt].l=last,ans[cnt].r=i;
s.clear();
last=i+1;
}
else
s.insert(a[i]);
}
if(!cnt)
{
printf("-1");
return 0;
}
ans[cnt].r=n;
printf("%d
",cnt);
for(int i=1;i<=cnt;i++)
printf("%d %d
",ans[i].l,ans[i].r);
return 0;
}