Description On the evening of 3 August 1492, Christopher Columbus departed from Palos de la Frontera with a few ships, starting a serious of voyages of finding a new route to India. As you know, just in those voyages, Columbus discovered the America…

Problem Description Peer-to-peer(P2P) computing technology has been widely used on the Internet to exchange data. A lot of P2P file sharing systems exist and gain much popularity in nowadays. Let's consider a simplified model of P2P file sharing syst…

Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. However, Ted is such a picky guy that in ev…

Description Facer is addicted to a game called "Tidy is learning to swim". But he finds it too easy. So he develops a new game called "Facer is learning to swim" which is more difficult and more interesting. In the new game, a robot na…

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the…

Description Students often have problems taking up seats. When two students want the same seat, a quarrel will probably begin. It will have very bad effect when such subjects occur on the BBS.  So, we urgently need a seat-taking-up rule. After severa…

HDU 6638 - Snowy Smile 题意 给你\(n\)个点的坐标\((x,\ y)\)和对应的权值\(w\),让你找到一个矩形,使这个矩阵里面点的权值总和最大. 思路 先离散化纵坐标\(y\)的值 对\(n\)个点根据横坐标\(s\)进行排序 枚举横坐标,按顺序把点扔到线段树里,以离散化后\(y\)的\(id\)为下标\(pos\),存到线段树里 因为线段树可以在\(\log{n}\)的时间内插入数值,在\(O(1)\)的时间内查询当前区间最大子段和(线段树区间合并) \(node[…

You live in a small town with R bidirectional roads connecting C crossings and you want to go from crossing 1 to crossing C as soon as possible. You can visit other crossings before arriving at crossing C, but it's not mandatory. You have exactly one…

Problem Description Josh Lyman is a gifted painter. One of his great works is a glass painting. He creates some well-designed lines on one side of a thick and polygonal glass, and renders it by some special dyes. The most fantastic thing is that it c…

标题效果 有着n巫妖.m精灵.k木.他们都有自己的位置坐标表示.冷却时间,树有覆盖范围. 假设某个巫妖攻击精灵的路线(他俩之间的连线)经过树的覆盖范围,表示精灵被树挡住巫妖攻击不到.求巫妖杀死所有精灵的时间.若无法所有杀死输出-1: 解题思路: 推断巫妖能否打到精灵用线段与点的最短距离来推断,若最短距离小于树的覆盖范围,就攻击不到. 最小时间能够跑费用流来解决,也能够二分图的最优匹配. 以下是代码: #include <set> #include <map> #include &l…