https://codeforces.com/contest/1311/problem/F 这是一道线段树类型的题: 可以用权值线段树或者树状数组来解: 所以,我们可以分为两部分,第一部分是计算出到当前点位置,小于等于当前点的速度的个数 ,总的个数乘当前点的速度 减去 小于等于当前点的速度的坐标总值即为答案: #include<bits/stdc++.h> using namespace std; typedef long long ll; ; int b[maxn]; struct node…
题:https://codeforces.com/contest/1311/problem/F 题意:给定x轴上的点以及他们的速度v,只在x轴上运动,求最小的dis之和,注意,这里的时间是可随意的,比如对于其中一个点 i 来说,只要其他点运动到离自己距离最小即可,而不是同步运动 分析:对于一对点 i 和 j 来说,以点 i 为基准,要是xj<xi,那么只要vj>xi就可以让dis为0,要是xj>xi,,那么只要vj<vi就可以让dis为0,其他情况就俩者不动就能保持dis最小为ab…
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 878 Accepted Submission(s): 353 Problem Description There are N points in total. Every point moves in certain direction and c…
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 710 Accepted Submission(s): 290 Problem Description There are N points in total. Every point moves in certain direction and…
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 964 Accepted Submission(s): 393 Problem Description There are N points in total. Every point moves in certain direction and c…
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 72 Accepted Submission(s): 18 Problem Description There are N points in total. Every point moves in certain direction and cer…
The Moving Points Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 612 Accepted Submission(s): 250 Problem Description There are N points in total. Every point moves in certain direction and…