HDU 4638-Group(线段树+离线处理)】的更多相关文章

题意: 给n个编号,m个查询每个查询l,r,求下标区间[l,r]中能分成标号连续的组数(一组内的标号是连续的) 分析: 我们认为初始,每个标号为一个组(线段树维护区间组数),从左向右扫序列,当前标号,要考虑和他相邻的标号的位置,若前面位置出现了和它相邻的标号, 则前面位置组数减一(因为可以合并成一组),查询区间离线处理,保证了查询的正确. #include <map> #include <set> #include <list> #include <cmath&g…
http://acm.hdu.edu.cn/showproblem.php?pid=4417 题意:找出给定区间内,有多少个数小于等于给定的数.用线段树维护的话会超时,要用到线段树的离线操作,对询问与数列都进行从小到大的排序,记录下标.从第一个询问开始,遍历数列,满足小于等于就插入到线段树中相应的位置.答案即为当前线段树中有多少个值.转换成了区间和,记录答案,最后一遍输出. 代码如下: #include<stdio.h> #include<algorithm> ; using na…
题意: 刚开始有一个空集合.有三种操作: 1.往集合中加入一个集合中不存在的数 x 2.从集合中删除一个已经存在的数 x 3.计算集合的digest sum并输出.       digest sum求法:将集合中所有数从小到大排序,得到a1<a2<...<an. digest sum = sum(ai) where i mod 5 = 3 数据范围: N ( 1 <= N <= 105 ) 1 <= x <= 109. For any "add x&qu…
根据题目意思,很容易得出,一个区间里面连续的段数即为最少的group数. 题解上面给的是用树状数组维护的. 询问一个区间的时候,可以一个一个的向里面添加,只需要判断a[i]-1 和 a[i]+1是否已经添加在内,如果两个都在,则总段数减1,如果两个都不在,总段数加1,其他情况总段数不变了.这里有一个需要深入理解的就是其实无论是按顺序添加还是随便添加,统计结果是不变的,但是要看怎么维护了. 每加入一个点,都会有一个改变量v[i],那么此时总段数就是sum{ v[i] } (1 <= i <= x…
Problem Description There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-1 are friends, Whose ID is i and i+1 are friends. These n men stand in line. Now we select an interval of men to make some group. K men in a group…
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两个题目都是求区间之内,不重复的数字之和,3333需要离散化处理................. 调试了一下午........说多了都是泪........... #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <string>…
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