poj 3320 尺取法

Description

Jessica’s a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

A very hard-working boy had manually indexed for her each page of Jessica’s text-book with what idea each page is about and thus made a big progress for his courtship. Here you come in to save your skin: given the index, help Jessica decide which contiguous part she should read. For convenience, each idea has been coded with an ID, which is a non-negative integer.

Input

The first line of input is an integer P (1 ≤ P ≤ 1000000), which is the number of pages of Jessica’s text-book. The second line contains P non-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type.

Output

Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book.

Sample Input

5
1 8 8 8 1
Sample Output

2

题解:

学习挑战程序设计。
经典尺取法。
先放到set去重得到知识点的不同数量。
利用一个map维护区间内的知识点数量，
从区间的最开头把s取出之后，页s上书写的知识点的出现次数就要减一，如果此时这个知识点的出现次数为0了，在同一个知识点再次出现前，不停将区间末尾t向后推进即可。每次在区间末尾追加页t时将页t上的知识点的出现次数加1，这样就完成了下一个区间上各个知识点出现次数的更新，通过重复这一操作可以以O(PlogP)的复杂度求出最小区间。

代码：

#include <iostream>
#include <map>
#include <cstdio>
#include <set>
using namespace std;
const int MAX_P =  1000000+100;
int P;
int a[MAX_P];

void solve()
{
   //计算全部知识点的总数
   set<int> all;
   for(int i=0;i<P;i++)
   {
       all.insert(a[i]);
   }
   int n = all.size(); //set可以去重

   //利用尺取法来求解
   int s=0,t=0,num=0;
   map<int,int> m;
   int res=P;
   for(;;)
   {
       while(t<P&&num<n)
       {
           if(m[a[t++]]++ == 0)
           {
               num++;//出现新的知识点
           }
       }
       if(num<n) break;
       res = min(res,t-s);
       if(--m[a[s++]]==0)
       {
           //某个知识点的出现次数为0
           num--;
       }
   }
   printf("%d\n",res);
}

int main()
{
    scanf("%d",&P);
    for(int i=0;i<P;i++)
        scanf("%d",&a[i]);
    solve();
    return 0;
}