A - Jessica's Reading Problem POJ - 3320 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…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13955   Accepted: 5896 Description A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) ar…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17562   Accepted: 6099 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 litt…

Jessica's Reading Problem 题目大意:Jessica期末考试临时抱佛脚想读一本书把知识点掌握,但是知识点很多,而且很多都是重复的,她想读最少的连续的页数把知识点全部掌握(知识点都在书上,每一页都是一个知识点) 这一题可以用3061的游标卡尺法,我们可以先数数书上倒到底有多少个知识点,因为知识点都是序数,我们可以用二分法直接找到O(PlogP),这里可以采用set模板直接偷懒了,然后我们就可以用游标卡尺法了,因为所有知识点都要出现一次,所以我们统计新的知识点的出现就好了,最…

题意:n页书,然后n个数表示各个知识点ai,然后,输出最小覆盖的页数. #include<iostream> #include<cstdio> #include<set> #include<map> using namespace std; ; int num[maxn]; int main(){ int n; set<int>ss; map<int, int>mm; scanf("%d", &n); ;…

Jessica's Reading Problem 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 ve…

1.POJ 3320 2.链接:http://poj.org/problem?id=3320 3.总结:尺取法,Hash,map标记 看书复习,p页书,一页有一个知识点,连续看求最少多少页看完所有知识点 必须说,STL够屌.. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio>…

题目链接: http://poj.org/problem?id=3320 题目大意:一本书有P页,每页有个知识点,知识点可以重复.问至少连续读几页,使得覆盖全部知识点. 解题思路: 知识点是有重复的,因此需要统计不重复元素个数,而且需要记录重复个数. 最好能及时O(1)反馈不重复的个数.那么毫无疑问,得使用Hash. 推荐使用map,既能Hash,也能记录对于每个key的个数. 尺取的思路: ①不停扩展R,并把扫过知识点丢到map里,直到map的size符合要求. ②更新结果. ②L++,map…

传送门:Problem 3320 参考资料: [1]:挑战程序设计竞赛 题意: 一本书有 P 页,每页都有个知识点a[i],知识点可能重复,求包含所有知识点的最少的页数. 题解: 相关说明: 设以a[start]开始的最初包含所有知识点的最少连续子序列为a[start,....,end]; mymap[ a[i] ] : 知识点 a[i] 在当前最少连续子序列中出现的次数. (1):求出所需复习的知识点总个数. (2):求出最先包含所有知识点的最少页数a[start,........,end].…

题目传送门 /* 尺取法:先求出不同知识点的总个数tot,然后以获得知识点的个数作为界限, 更新最小值 */ #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MA…