Problem Description Pony is the boss of a courier company. The company needs to deliver packages to n offices numbered from 1 to n. Especially, the s-th office is the transfer station of the courier company. There are x ordinary two-way roads and y o…

http://www.lydsy.com/JudgeOnline/problem.php?id=2100 这题我要吐血啊 我交了不下10次tle.. 噗 果然是写挫了. 一开始没加spfa优化果断t 然后看了题解加了(加错了T_T)还是tle..我就怀疑数据了... 噗 原来我有个地方打错了.. 这个spfa的队列优化真神.. #include <cstdio> #include <cstring> using namespace std; #define rep(i, n) fo…

Delivery Route 题目:有n个派送点,x条双向边,y条单向边,出发点是s,双向边的权值均为正,单向边的权值可以为负数,对于单向边给出了一个限制:如果u->v成立,则v->u一定不成立.问,从s出发,到其他所有点的最短路是多少(包括s). 思路:对于单向边的限制,我们可以这么理解:双向边相连接的点一定组成一个强连通分量,如果一条单向边存在于某个强连通分量中,可以得出:如果“u -> v”,则一定“v -> u”,可以推出单向边一定只存在于连接两个强连通分量,且还可以推出,…

It is my great honour to introduce myself to you here. My name is Aloysius Benjy Cobweb Dartagnan Egbert Felix Gaspar Humbert Ignatius Jayden Kasper Leroy Maximilian. As a storyteller, today I decide to tell you and others a story about the hero Huri…

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1496    Accepted Submission(s): 723 Problem Description Kelukin is a businessman. Every day, he travels arou…

Problem DescriptionOne day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki…

用到了网络流的思想(大概).新建一个源点s,所有边权扩大两倍,然后所有的点向s连边权为点权的无向边,然后以s为起点跑spfa(S什么L优化的),这样每个点到s的距离就是答案. 原因的话,考虑答案应该是min(2*dis[i][j]+a[j]} ),那么每个点到s的距离就是若干条边边权的二倍加上某个点的点权,并且这个组合是最小的.证毕. #include<iostream> #include<cstdio> #include<queue> using namespace…

洛谷数据好强啊,普通spfa开o2都过不了,要加双端队列优化 因为是双向边,所以dis(u,v)=dis(v,u),所以分别以pa1和pa2为起点spfa一遍,表示pb-->pa1-->pa2和pb-->pa2-->pa1,取个min即可 #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=200005,inf=2e9+7; in…

Rate Distortion Optimization 搜索时,一个不可避免的问题就是如何对mv进行比较,从而得到最优 对于同一压缩算法来说,码率越高表示图像质量越好.失真越小,但是码率越高要求更大的存储空间,也会增加网络传输的压力.因此在码率与失真中找出平衡点,使压缩效果最优,这种方法叫做R-D Optimization(码率失真优化) 典型的高码率下,码率与失真关系(R-D关系式)为 $R(D) = {\alpha}{ln(\frac{\delta^2}{D})}$ $R$为码率,$\al…

用双端队列,当待加入元素小于队首元素时 加入队首, 否则 加入队尾 slf优化 if(!vis[e.v]) { if(Q.empty()) Q.push_front(e.v); else { if(d[e.v] < d[Q.front()]) Q.push_front(e.v); else Q.push_back(e.v); } vis[e.v] = ; } next从结构体中分离出来 用数组记录 using namespace std; , INF = 0x7fffffff; int n, m…