However, TT is a nice and lovely girl. She doesn't have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed. What's more, if FF finds an answer to be wrong, he will ignore it when judging next answers....
HDU - 3038 How Many Answers Are Wrong (带权并查集) 题意:有长度为N的序列A,每个值都位置。给出M条信息:A[L...R]之和为S。但是不一定每一条都是正确的,求出错误的信息个数。 分析:经典的带权并查集问题。一个区间内的和可以转化为右端点到其父亲结点的距离,而父亲节点是L-1。如果R点和L-1点本身...
However, TT is a nice and lovely girl. She doesn't have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed. What's more, if FF finds an answer to be wrong, he will ignore it when judging next answers....
2Author: Leo.W 3Descriptipn: 。 4How to Do: sum[x]表示由第x+1个元素到根元素的元素之和。三个关键点: 51、merge(a-1,b,c);a-1而不是a,因为按我们的思路c=[b+1]-[(a-1)+1]得到的,正好是a~b; 62、sum[px]=sum[y]-sum[x]+c;(注意此处c=元素x+1到元素y的和) 73、路径压缩时...
HDU 3038 How Many Answers Are Wrong(带权并查集) 题目大意就是有M个数,不知道它们具体的值,但是知道某两个数之间(包括这两个数)的所有数之和,现在给出N个这样的区间和信息,需要判断有多少个这样的区间和与前边已知的区间和存在矛盾。例如给出区间和[1,4]为20,[3,4]为15,再给出[1,2]为30,显然这个...
链接:https://vjudge.net/problem/HDU-3038 思路:带权并查集,首先我们要考虑在什么情况下会出错,当且仅当某个区间开头和位置以及和都确定并且产生矛盾的时候,于是我们建立一个带权并查集,每个区间查询其左端点-1的节点(因为左端点也要算在和内)与右端点的节点的祖先节点,如果相同说明通过其他的操作已经可以推算出...
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一、内容 TT and FF are ... friends. Uh... very very good friends -___-bFF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).T
3038How Many Answers Are Wrong 2818Building Block 3234Exclusive-OR 2586How far away? 2874Connections between cities 3486Interviewe 2888Check Corners 【矩阵乘法】 1575Tr A 1757A Simple Math Problem 2254奥运 2256Problem of Precision 1558Gauss Fibonacci 2604Queuing 2276Kiki & Little Kiki 2 2855Fibonacci...
HDU 3038 How Many Answers Are Wrong(带权并查集,真的很难想到是个并查集!!!)... How Many Answers Are Wrong Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14546 Accepted Submission... ...