| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23144 | Accepted: 7819 |
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
Source
sa
poj 3320
题意：
第i页有知识点a[i]，求学完所有的知识点的最少连续页数
共有p个数字，求连续的几个数字，这几个数字包含所有输入中出现过的数字
求这样的数字序列的最小值，用尺取法去做。
做法：
假设从某一页s开始阅读，为了覆盖所有的知识点读到t页，这样的话如果从s+1开始阅读，那么必须读到t'>=t位置，尺取法。
如果一个区间的子区间满足条件，那么在区间推进到该处时，右端点会固定，左端点会向右移动到其子区间，且其子区间会是更短的，只是需要存储所选取的区间的知识点的数量，那么使用map进行映射以快速判断是否所选取的页数是否覆盖了所有的知识点
Source Code
| Problem: 3320 | User: bbsh | |
| Memory: 1908K | Time: 422MS | |
| Language: G++ | Result: Accepted |
- Source Code
#pragma GCC optimize(3)
#pragma G++ optimize(3)
#include<map>
#include<set>
#include<stdio.h>
using namespace std;
set<int>s;
map<int,int> cnt;
const int N=1e6+5;
int n,p,ans,st,en,sum,a[N];
int main(){
scanf("%d",&p);
for(int i=0;i<p;i++) scanf("%d",a+i),s.insert(a[i]);
n=s.size();
for(ans=p;;){
while(en<p&&sum<n) if(cnt[a[en++]]++==0) ++sum;
if(sum<n) break;
ans=min(ans,en-st);
if(--cnt[a[st++]]==0) --sum;
}
printf("%d
",ans);
return 0;
}
更多例题请参考这里