垃圾poj交不上去 /* 按权值从小到大排序, 两个树状数组维护权值小于等于并且在i左边的点的个数和权值 */ #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define maxn 20005 #define ll long long struct node{ int w,x; bool operator<(…

题目链接: https://vjudge.net/problem/POJ-1990 题目大意: 一群牛参加完牛的节日后都有了不同程度的耳聋,第i头牛听见别人的讲话,别人的音量必须大于v[i],当两头牛i,j交流的时候,交流的最小声音为max{v[i],v[j]}*他们之间的距离.现在有n头牛,求他们之间两两交流最少要的音量和. 解题思路: 使用树状数组,首先将二元组按照v的大小从小到大排序,这样可以保证每头牛比前面的牛的v大,计算它和它前面牛的音量和的时候,就可以直接用该头牛的v,还需要计算出|…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3015 题目大意: 有一些树,这些树的高度和位置给出.现在高度和位置都按从小到大排序,对应一个新的rank,任意两棵树的值为min(高度的rank) * abs(位置差的绝对值).问所有任意两棵树的值的和是多少. 解题思路: 按照题意离散化,然后对H从大到小排序,这样可以保证前面的树高度都比当前的高(或者相等).在计算的时候就可以使用当前的H. 和POJ-1990类似 #include<iostre…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3015 题意:给定n组数,每组数有值x和值h,求n组数两两的val的总和.将所有x和所有h分别离散化(不去重)变成x'和h',val(i,j)为abs(x'i-x'j)*min(hi',hj'). 如: x,    h——>x',h' 10,100——>1,1 50,500——>4,4 20,200——>3,3 20,100——>1,1 思路:只要把n*n优化成n*logn就可以过…

Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of cours…

都是一类题目,推导调试比较烦,想出来还是不难的 /* 给定n个点对,按一维升序排序一次,每个点的序号为Di,按二维升序排序一次,每个点的序号为Hi 求sum{w(i,j)} w(i,j)=abs(Di-Dj)*min(Hi-Hj) 那么将所有的点按照H升序排列,两个树状数组,一个维护区间d的个数,一个维护区间d的总和 每个点的贡献值=树状数组中D小于其的+D大于其的abs 按升序遍历每个点后,将该点在树状数组中删掉 */ #include<bits/stdc++.h> using namesp…

Problem Description You have a sequence {a1,a2,...,an} and you can delete a contiguous subsequence of length m. So what is the minimum number of inversions after the deletion.   Input There are multiple test cases. The first line of input contains an…

一共6种情况,a < b且Aa < Ab, c < d 且Ac > Ad,这两种情况数量相乘,再减去a = c, a = d, b = c, b = d这四种情况,使用树状数组维护,le[i]表示i左边比他小的数数量,le1[i]表示i左边比他大的数数量,ri[i]表示i右边比他小的数数量,ri1[i]表示i右边比他大的数数量. 跑两次树状数组,求出这四个数组值. 注意处理数值相等的情况. #include<cstdio> #include<iostream>…

                                                                    MooFest Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7077   Accepted: 3181 Description Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest"…

题目链接: Inversion Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1087    Accepted Submission(s): 323 Problem Description You have a sequence {a1,a2,...,an} and you can delete a contiguous subsequ…