模拟,注意需要比较队头与当前时间的大小关系. #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 10005 #define INF 0xffffff typedef struct { int i, t; } node_t; node_t Q[][MAXN]; int buf[MAXN]; ], front[]; int main() { int case_n; int n, t, m…

Dinic不同实现的效率果然不同啊. /* 3277 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #include <…

Ferry Loading IIITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 463 Accepted Submission(s): 110 Problem DescriptionBefore bridges were common, ferries were used to transport cars across rivers. Ri…

[LeetCode]732. My Calendar III解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/my-calendar-iii/description/ 题目描述: Implement a MyCalendarThree class to store your events. A new event can always be added. Your class will have one method, book…

Difficulty:easy  More:[目录]LeetCode Java实现 Description Design and implement a TwoSum class. It should support the following operations: addand find.add(input) – Add the number input to an internal data structure.find(value) – Find if there exists any…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

错误信息: /usr/local/memcacheq/bin/memcacheq: error while loading shared libraries: libevent-2.0.so.5: cannot open shared object file: No such file or directory   查找链接包 : LD_DEBUG=libs /usr/local/memcacheq/bin/memcacheq -v | grep libevent      27382: fin…

在已有的Control.Loading控件基础上结合CSS3 animation属性实现 .nz-loading .nz-loader { display: block; -webkit-animation: shift-rightwards 3s ease-in-out infinite; -moz-animation: shift-rightwards 3s ease-in-out infinite; -ms-animation: shift-rightwards 3s ease-in-ou…

http://www.lydsy.com/JudgeOnline/problem.php?id=1406 题意:求$0<=x<n, 1<=n<=2,000,000,000, 且x^2 \equiv 1 \pmod{n}$的所有$x$ #include <bits/stdc++.h> using namespace std; typedef long long ll; set<ll> s; int main() { ll n; scanf("%lld…

DP/四边形不等式 裸题环形石子合并…… 拆环为链即可 //HDOJ 3506 #include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define…