终于又回到熟悉的Round了 数学 A - Pouring Rain 设个未知数,解方程,还好没有hack点 #include <bits/stdc++.h> typedef long long ll; const int N = 1e5 + 5; const double PI = acos (-1.0); int main() { double d, h, v, e; scanf ("%lf%lf%lf%lf", &d, &h, &v, &…

题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一个满足条件的四个点. 题解: 首先预处理出任意两点的最短距离,用队列优化的spfa跑:O(n*n*logn) 现依次访问四个点:v1,v2,v3,v4 我们可以枚举v2,v3,然后求出v2的最远点v1,v3的最远点v4,为了保证这四个点的不同,直接用最远点会错,v1,v4相同时还要考虑次最远点来替换…

题目链接:http://codeforces.com/contest/667/problem/D 给你一个有向图,dis[i][j]表示i到j的最短路,让你求dis[u][i] + dis[i][j] + dis[j][v]的最大值,其中u i j v互不相同. 先用优先队列的dijkstra预处理出i到j的最短距离(n^2 logn).(spfa也可以做) 然后枚举4个点的中间两个点i j,然后枚举与i相连节点的最短路dis[u][i](只要枚举最长的3个就行了),接着枚举与j相连节点的最短路…

B. World Tour 题目连接: http://www.codeforces.com/contest/666/problem/B Description A famous sculptor Cicasso goes to a world tour! Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he…

A. Reberland Linguistics 题目连接: http://www.codeforces.com/contest/666/problem/A Description First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The educati…

题目链接: 题目 A. Reberland Linguistics time limit per test:1 second memory limit per test:256 megabytes 问题描述 First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university…

C. Reberland Linguistics time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talent…

题意:给一个初始串s,和m个模式串,q次查询每次问你第l到第r个模式串中包含\(s_l-s_r\)子串的最大数量是多少 题解:把初始串和模式串用分隔符间隔然后建sam,我们需要找到在sam中表示\(s_l-s_r\)子串的状态节点(先找到\(s_r\)对应的节点,然后倍增parent树即可),我们需要找到包含\(s_l-s_r\)的模式串,这些节点肯定在parent树上位于我们找到的状态节点的子树上.那么我们按sam的topo序进行线段树合并,线段树区间表示l,r模式串中最大匹配值以及下标.每次…

第一题直接算就行了为了追求手速忘了输出yes导致wa了一发... 第二题技巧题,直接sort,然后把最大的和其他的相减就是构成一条直线,为了满足条件就+1 #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<iomanip> #…

D. World Tour   A famous sculptor Cicasso goes to a world tour! Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will en…