I. Move Between Numbers   time limit per test 2.0 s memory limit per test 256 MB input standard input output standard output You are given n magical numbers a1, a2, ..., an, such that the length of each of these numbers is 20 digits. You can move fro…

畅通project续 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28578    Accepted Submission(s): 10382 Problem Description 某省自从实行了非常多年的畅通project计划后.最终修建了非常多路. 只是路多了也不好,每次要从一个城镇到还有一个城镇时.都有很多种道路方案能够选择…

不写普通模板了,还是需要优先队列优化的昂 #include<stdio.h> //基本需要的头文件 #include<string.h> #include<queue> #include<algorithm> #include<vector> using namespace std; typedef pair<int,int> pii; const int INF=0x3f3f3f3f; ; ; ],val[maxm<<]…

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer John's field ha…

http://acm.hdu.edu.cn/showproblem.php?pid=2544 #include<iostream> #include<queue> #include<functional> #include<algorithm> #include<cstring> #include<vector> using namespace std; ; const int INF = 1e7; typedef pair<i…

1.裸题 hdu2544 http://acm.hdu.edu.cn/showproblem.php?pid=2544 Way1: 好像不对 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <stdbool.h> #include <set> #include <vector> #include <ma…

ROADS Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12436 Accepted: 4591 Description N cities named with numbers 1 - N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that…

传送门:地铁 思路:拆点,最短路:拆点比较复杂,所以对边进行最短路,spfa会tle,所以改用Dijkstra(优先队列优化) 模板 /************************************************************** Problem: User: youmi Language: C++ Result: Accepted Time: Memory: *****************************************************…

一下午都在学最短路dijkstra算法,总算是优化到了我能达到的水平的最快水准,然后列举一下我的优化历史,顺便总结总结 最朴素算法: 邻接矩阵存边+贪心||dp思想,几乎纯暴力,luoguTLE+MLE, 算优点??:但好写好想,比下面的代码短了差不多一半. #include <iostream> #include <cstdio> #include <algorithm> using namespace std; int main() { ][],point,i,j,…

题意: n个点,m条边,问从1走到n的最短路,其中有K次机会可以让一条路的权值变成0.1≤N≤10000;1≤M≤500000;1≤K≤20 题解: 拆点,一个点拆成K个,分别表示到了这个点时还有多少次机会.(x,k)-->(y,k-1),cost=0 或 (x,k)-->(y,k),cost=a[i].d;这题数据比较大, 需要很多优化.(应该只是蒟蒻我才需要这么多优化..)1.不用spfa(时间复杂度不稳定),用dijkstra+优先队列优化2.拆点不拆边.g[i]表示i这个点是由谁拆分出…