556D - Case of Fugitive 思路:将桥长度放进二叉搜索树中(multiset),相邻两岛距离按上限排序,然后二分查找桥长度匹配并删除. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; struct node { ll x; ll y; int id; bool operator <(const node &a)const { return y<a.y; } }…

Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water. The only dry land there is an archipelago of n narrow islands located in a row. For more comfort l…

题目连接: http://codeforces.com/problemset/problem/555/B 题目大意: 有n个岛屿(岛屿在一列上,可以看做是线性的,用来描述岛屿位置的是起点与终点),m个桥,给出每个岛屿的位置和桥的长度,问是否可以把n个岛屿连起来? 解题思路: 排序+贪心,对于n个岛屿,相邻的两个之间起点和端点可以转化为n-1个连接桥的长度区间,把区间升序排列. 对于m个桥升序排列,对于每一个桥枚举每个可以合法覆盖的区间,选取最优的,选取的时候满足L<bridge_length<…

题意:有n-1个缝隙,在上面搭桥,每个缝隙有个ll,rr值,ll<=长度<=rr的才能搭上去.求一种搭桥组合. 经典问题,应列入acm必背300题中.属于那种不可能自己想得出来的题.将二元组[ll,rr]排序(ll相同时再rr),长度x排序(升序).一个全局优先队列pq(rr小的顶部).for循环,对每个x,将ll比它小的放入优先队列pq,如果pq仍为空,说明这块桥用不上,不为空,看top的rr是否大于x,如果大于,这块桥就能用上,并且给当前的top一定是可行的. 乱码: #pragma co…

B. Case of Fugitive Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/B Description Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost ful…

\(>Codeforces \space 555 B. Case of Fugitive<\) 题目大意 : 有 \(n\) 个岛屿有序排列在一条线上,第 \(i\) 个岛屿的左端点为 \(l_i\) 右端点为 \(r_i\) ,岛屿之间两两不相交, 现在对于每一个 \(1 \leq i < n\) 第 \(i\) 岛屿要和第 \(i + 1\) 岛屿之间建一座桥,桥的长度左右端点必须得在岛上.现在有 \(m\) 座已经长度建好的桥梁,试找出一种岛屿和桥匹配的方案,使得任意两座岛屿之间的…

B. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet a…

D. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet a…

[Codeforces 555E]Case of Computer Network(Tarjan求边-双连通分量+树上差分) 题面 给出一个无向图,以及q条有向路径.问是否存在一种给边定向的方案,使得这q条路径都能被满足.(如果有一条边是从a->b),而经过它的路径是从b->a,那么久不满足).只需要判断,不用输出方案. 分析 对于一个有向环,显然它可以允许各个方向的路径通过.所以我们只要把无向图里的边-双联通分量建成环,然后就不用考虑了.影响答案的只有桥. 所以我们求出所有桥,然后缩点,把图…

传送门 D. Restructuring Company time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Even the most successful company can go through a crisis period when you have to make a hard decision — to rest…