给两组字符串,最多有多少对相同. map做映射判断一下. #include <iostream> #include <cstdio> #include <map> #include <string> using namespace std; map<string,int> mp; string s; int n,ans; int main() { while(~scanf("%d",&n)) { ans=; mp.cl…

Judging Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=114147 Description The NWERC organisers have decided that they want to improve the automatic grading of the submissions for the contest,…

-----------------------前面的两场感觉质量不高,就没写题解----------------------------- A .Around the Track pro:给定内多边形A和外多边形B,求最短路径,蛮子路径再A之外,B之内. sol:如果没有B,就是求凸包,有了B,我们在做凸包的时候,有形如“a-b-c,b在内部,删去b,连接a-c的操作”,如果a-c和B不相交,直接删去b,否则用B的凸包代替b处. #include <bits/stdc++.h> using n…

B:Biking Duck 题意:现在有一个人要从(x1,y1)点走到(x2,y2)点, 现在走路的速度为v. 还有骑自行车的速度v2,自行车要从某个自行车站到另一个自行车站,现在我们可以视地图的边界都为自行车站,求最小时间是多少. 题解:对于忽视边界的问题,那就是暴力枚举.现在有了边界, 我们假设一个点在边界,那么我们可以枚举另一个自行车点之后,在某个边界上三分另一个点. 下一个情况就是2个自行车站都在边界上,我们三分套三分去维护找到最小值. 思维不怎么麻烦,就是写的有点麻烦. 代码: #in…

Hanoi Tower Troubles Again! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 602    Accepted Submission(s): 418 Problem Description People stopped moving discs from peg to peg after they know the…

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…

UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…

UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…

链接:ZOJ1239 Hanoi Tower Troubles Again! Description People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with…

题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离: 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath&…