题目出处:http://codeforces.com/problemset/problem/886/C 题目大意:很多墓穴之间有通道,探险家来回穿梭并记录日志 日志规则:第一次到该墓穴计时间t,0<=t<当前时间i:再次经过记录i #include<iostream> #include<set> using namespace std; //集合的运用 int main(){ set<int> all; int n,x; cin>>n; ;i&l…

突然发现百度不到这题的单独题解(果然是因为这是水题么),那我就来写一个了~ 先把题给贴了. C. Petya and Catacombs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A very brave explorer Petya once decided to explore Paris catacombs. Sin…

C. Petya and Catacombstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputA very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his exploration is…

C. Petya and Catacombs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his explor…

A very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his exploration is just walking through the catacombs. Catacombs consist of several rooms and bidirectional passages between some pairs of the…

题目描述 A very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his exploration is just walking through the catacombs. Catacombs consist of several rooms and bidirectional passages between some pairs o…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 看看时间戳为i的点有哪些. 每次优先用已经访问过的点. 如果不行就新创一个点. 注意新创点的时间戳也是i. [代码] #include <bits/stdc++.h> using namespace std; const int N = 2e5; int n; int now = 1, tot = 1; vector <int> have[N + 10]; int before[N + 10]; int main(…

A. Petya and Catacombs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A very brave explorer Petya once decided to explore Paris catacombs. Since Petya is not really experienced, his explor…

ACM ICPC 每个队伍必须是3个人 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<vector> #include<algorithm> using namespace std; int cmp(const void * x, const void * y) { //x < y : -; } ]; int main() { #ifndef ONLINE_…

[BZOJ2246][SDOI2011]迷宫探险(搜索,动态规划) 题面 BZOJ 洛谷 题解 乍一看似乎是可以求出每个东西是陷阱的概率,然而会发现前面走过的陷阱是不是陷阱实际上是会对当前状态产生影响的.考虑一下状压,因为出了是陷阱和不是陷阱,还有一种情况是未知.所以三进制状压. \(0\)表示是有害陷阱,\(1\)表示不是,\(2\)表示未知. 那么假如我们知道了一个当前的三进制状态,如何确定当前的某个未知的陷阱是否有害的概率呢? 这个显然可以暴力提前预处理出来. 那么这就很好办了,设\(f[…