注意题目保证不会有一个矩形完全包括另一个矩形的情况 时间序上从后往前看,一个坐标\((x,y)\)加进来之前,如果已经有\(x_i<x\),则对结果的贡献为\(x-x_i\);若不存在\(x_i<x\),则对结果的贡献为\(x\).纵坐标\(y\)同理. 用两个集合维护已经存在的横纵坐标,每次用set自带的lower_bound查找第一个大于等于当前x和y的值,然后将迭代器--以查找上一个元素(即第一个小于它的元素). #include<bits/stdc++.h> #define…

ACM-ICPC 2018 徐州赛区网络预赛 G. Trace (思维,贪心) Trace 问答问题反馈 只看题面 35.78% 1000ms 262144K There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx ,…

https://nanti.jisuanke.com/t/31459 题意 n个矩阵,不存在包含,矩阵左下角都在(0,0),给右上角坐标,后来的矩阵会覆盖前面的矩阵,求矩阵周长. 分析 set按照x或者y从大到小排序,从后往前遍历,放入set,找到当前矩阵在set中的位置,当前矩阵的x或者y,与set中后一位矩阵的x或者y的差值,就是增加的横线或者竖线的长度 感觉套个线段树求周长的模板也行. #include<queue> #include<cstring> #include<…

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of…

https://nanti.jisuanke.com/t/31459 思路 凡是后面的轨迹对前面的轨迹有影响的,可以尝试从后往前扫 区间修改需要push_down,单点更新所以不需要push_up(用于区间查询) 多颗线段树的时候将函数写进结构体里,这样所有函数只需要写一次了 #include<bits/stdc++.h> #define pb push_back #define M 100005 using namespace std; int n,i,x[M],y[M],sz,tp; lo…

任意门:https://nanti.jisuanke.com/t/31459 There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every ti…

Problem:Portal传送门 Problem:Portal传送门  原题目描述在最下面.  我理解的题意大概是:有n次涨潮和退潮,每次的范围是个x×y的矩形,求n次涨退潮后,潮水痕迹的长度.  不存在此i,j∈[1,n],i≠j,xi≤xj且yi≤y Solution:  每次潮水可能会冲刷掉之前的潮水的一部分痕迹,但是它又不会完全冲刷.  考虑从最后一次潮水开始往前算贡献,分x,y方向计算.  假设此次潮水的范围是[xi,yi],找出大于等于xi的最大Y,则y方向的新增痕迹长度为yi−Y…

赛后和队友讨论了一波,感谢无敌的队友给我细心的讲题 先埋坑 #include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> using namespace std; ; ; int treex[maxx]; int treey[maxx]; ; int lowbit(int x){ return x&-x; } void add(int i,int x,int *…

题目链接: https://nanti.jisuanke.com/t/31459 具体思路: 先顺序输入,然后回溯,假设已经加入了n个点,那么在加入的同时,首先看一下原先x轴上已经有过的点,找到第一个最接近第n个点并且小于第n个点的坐标,y轴同理.如果没有找到,比如说第一个,那么他减去的就是0.然后再就是代码,如果用数组存储的话,每进入一个点都需要对原来的数组进行重新排序,然后再去查找所需的值,这样会超时.可以用set存储,每进入一个数都会进行重新排序,然后存储完之后,再对set进行二分查找,找…

ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer J. Maze Designer After the long vacation, the maze designer master has to do his job. A tour company gives him a map which is a rectangle. The map consists of N \times MN×M little squares. That is to say, the h…

ACM-ICPC 2018 徐州赛区网络预赛 A.Hard to prepare 枚举第一个选的,接下来的那个不能取前一个的取反 \(DP[i][0]\)表示选和第一个相同的 \(DP[i][1]\)表示选和第一个取反的 \(DP[i][2]\)表示选其他的 状态转移方程直接看代码好了 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #inc…

H.Ryuji doesn't want to study 27.34% 1000ms 262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That…

传送门 题面: In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to consta…

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly fac…

262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That means, if he reads books from ll to rr, he…

262144K   Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <x, y>. If xi​= xj​ and yi…

Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, he calls Kimura KMR, and calls Suzuki YJSNPI. One day he read a book about SHA-256,which can transit a string into just 256 b…

链接 https://nanti.jisuanke.com/t/31456 参考题解  https://blog.csdn.net/ftx456789/article/details/82590044 #include <bits/stdc++.h> #define pb push_back #define mp make_pair #define fi first #define se second #define all(a) (a).begin(), (a).end() #define…

G. Lpl and Energy-saving Lamps 42.07% 1000ms 65536K   During tea-drinking, princess, amongst other things, asked why has such a good-natured and cute Dragon imprisoned Lpl in the Castle? Dragon smiled enigmatically and answered that it is a big secre…

There are N children in kindergarten. Miss Li bought them NNN candies. To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1...N), and each time a child is invited, Miss Li…

There are NN children in kindergarten. Miss Li bought them NN candies. To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1...N)(1...N), and each time a child is invited,…

ACM-ICPC 2018 徐州赛区网络赛  去年博客记录过这场比赛经历:该死的水题  一年过去了,不被水题卡了,但难题也没多做几道.水平微微有点长进.     D. Easy Math 题意:   给定 \(n\), \(m\) ,求 \(\sum _{i=1}^{m} \mu(in)\) .其中 $ 1 \le n \le 1e12$ , $ 1 \le m \le 2e9$ ,\(\mu(n)\) 为莫比乌斯函数.   思路:   容易知道,\(i\) 与 \(n\) 不互质时, \(\m…

A. Hard to prepare 题意:有n个客人做成一圈,有$2^k$种面具,对于每种面具有一种面具不能使相邻的两个人戴,共有多少种做法. 思路: 把题意转化成相邻的人不能带同种面具.总数为$(2^k)^n$,减去一对相邻的客人戴同种面具$(2^k)^{(n-1)}*C(n,1)$,其中重复了两对相邻的客人戴同种面具$(2^k)^{(n-2)}*C(n,2)$,依次容斥. 最后所有人都戴同种面具的情况额外考虑,当n是奇数时,n-1对客人相同即所有人相同.n为偶数时,n-1对客人相同时用公式…

A. Hard to prepare #include <bits/stdc++.h> using namespace std; ; ]; ]; int main() { int T; int n,k; bit[] = ; ; i <= ; i++) { bit[i] = *bit[i-]%MOD; } scanf("%d", &T); while (T--) { scanf("%d%d", &n, &k); f[] = b…

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of…

https://nanti.jisuanke.com/t/31462 题意 一个N*M的矩形,每个格点到其邻近点的边有其权值,需要构建出一个迷宫,使得构建迷宫的边权之和最小,之后Q次查询,每次给出两点坐标,给出两点之间的最短路径 分析 可以把每个格点视作视作图的点,隔开两点的边视作图的边,则构建迷宫可以视作求其生成树,剩余的边就是组成迷宫的墙.因为要花费最小,所以使删去的墙权置最大即可,呢么就是求最大生成树即可.然后每次查询相当于查这个最大生成树上任意两点的最短距离,到这一步就是个LCA了. 这…

https://nanti.jisuanke.com/t/31462 要求在一个矩形中任意选两个点都有唯一的通路,所以不会建多余的墙. 要求满足上述情况下,建墙的费用最小.理解题意后容易想到首先假设全部墙都建起来,然后拆掉费用最大的边使图成为一棵树,就是求一颗最大生成树 求出最大生成树后,求任意两点的距离,直接用lca就可以 思路 #include<bits/stdc++.h> #define M 300005 #define pb push_back using namespace std;…

链接https://nanti.jisuanke.com/t/31455 思路 首先先枚举把剩下的数填入星号的情况(其实就是枚举星号的排列),这是对方所能知道的所有信息,然后对方将取八种决策中最优的情况,而因为井号的存在,所以其排列也会影响每种决策的分数,所以接着要枚举井号的排列情况,对于每种情况累加每个决策的分数,最后枚举完后,要除以井号排列数(期望=分数*概率),然后对方便会选择期望最高的决策,累加到最后答案中,枚举完所有星号的之后,需要将最后答案除以星号排列数 #include<bits/…

G. Give Candies There are N children in kindergarten. Miss Li bought them N candies. To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1...N) and each time a child is inv…

Morgana is playing a game called End Fantasy VIX. In this game, characters have nn skills, every skill has its damage. And using skill has special condition. Briefly speaking, if this time you use skill "x", then next time you can use skill &quo…