Disharmony Trees Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.…

/* 写完这篇博客有很多感慨,过去一段时间都是看完题解刷题,刷题,看会题解,没有了大一那个时候什么都不会的时候刷题的感觉,这个题做了一天半,从开始到结束都是从头开始自己构思的很有感觉,找回到当初的感觉 */ Disharmony Trees Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 90 Accepted Submission(s):…

Disharmony Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 663    Accepted Submission(s): 307 Problem Description One day Sophia finds a very big square. There are n trees in the square. T…

Disharmony Trees HDU - 3015 One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them. She finds that trees maybe disharmony and the Disharmony Value between two trees is associa…

Problem Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them. She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with…

Problem Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with…

题意:给你n棵树,每棵树上有两个权值X H 对于X离散化 :3 7 1 5 3 6 -> 2 6 1 4 2 5,对于H一样 然后F = abs(X1-X2)   S=min(H1,H2) 求出每一对F*S的总和 可以看到一边是求每个数与其他数的最小值,一边是求每个数与其他数的差距.因此我们可以排序一边,处理另一边. 我们排序H,因为这样对于固定一个Xi Hi,从小到大每次都是Hi去乘以Xi与剩下的所有X的差的总和. 这样我们就可以使用树状数组维护两个值:每个位置值的个数,每个位置值的总大小,接…

题解:在路边有一行树,给出它们的坐标和高度,先按X坐标排序.记录排名,记为rankx,再按它们的高度排序,记录排名,记为rankh.两颗树i,j的差异度为 fabs(rankx[i]-rankx[j])*min(rankh[i],rankh[j]) 最后求出任异两颗树差异度的和. 题解:首先,需要解决的是min(rh)的问题,对于这个问题,只要离散化之后按rh从大到小排序顺序求解即可,然后用树状数组维护之前出现rx比当前值小的个数与总和,那么同时就可以计算rx比当前大的个数和总和了,那么就可以依…

题意:给出n棵树,给出横坐标x,还有它们的高度h,先按照横坐标排序,则它们的横坐标记为xx, 再按照它们的高度排序,记为hh 两颗树的差异度为 abs(xx[i] - xx[j]) * min(hh[i],hh[j]),求所有的差异度的和 和上面一道题一样,只不过这题是要Min的值,就将h从大到小排序,保证每一个h都是当前最小的 然后维护比当前x小的坐标的个数,当前区间的总和,前面比x小的坐标的和 #include<iostream> #include<cstdio> #inclu…

#include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; struct node { ll x,h; } Tr[]; ],tr2[]; bool cmp(node s1,node s2) { return s1.x<s2.x; } bool cmp1(node s1,node s2) { return s1.h<s2.h;…