Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a little hook on the floor in the center of every fr…

Labyrinth Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 4004   Accepted: 1504 Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them eithe…

Labyrinth 题目链接: http://acm.hust.edu.cn/vjudge/contest/130510#problem/E Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or f…

题目连接 http://poj.org/problem?id=1383 Labyrinth Description The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a litt…

树的直径:树上的最长简单路径. 求解的方法是bfs或者dfs.先找任意一点,bfs或者dfs找出离他最远的那个点,那么这个点一定是该树直径的一个端点,记录下该端点,继续bfs或者dfs出来离他最远的一个点,那么这两个点就是他的直径的短点,距离就是路径长度.具体证明见http://www.cnblogs.com/wuyiqi/archive/2012/04/08/2437424.html 其实这个自己画画图也能理解. POJ 1985 题意:直接让求最长路径. 可以用dfs也可以用bfs bfs代…

题目链接:http://poj.org/problem?id=2299 题目大意:给定n个数,要求这些数构成的逆序对的个数. 可以采用归并排序,也可以使用树状数组 可以把数一个个插入到树状数组中, 每插入一个数, 统计比他小的数的个数,对应的逆序为 i- getsum( data[i] ),其中 i 为当前已经插入的数的个数, getsum( data[i] )为比 data[i] 小的数的个数,i- getsum( data[i] ) 即比 data[i] 大的个数, 即逆序的个数.最后需要把…

POJ - 3067 题意:有(1-n)个城市自上到下在左边, 另有(1-m)个城市自上到下在右边,共有m条高速公路,现求这m条直线的交点个数,交点不包括在城市处相交. 题解:先将高速公路读入,然后按照左边城市序号小的在前,左边序号相同的按照右边城市序号小的在前,然后根据右边的城市序号求逆序数就可以解决了. 因为当处理一条边的时候,这条边一定会和左边序号小于它的且右边序号大于它的边有一个交点. 代码: 没想到cin, cout 关了同步流也T了, 然后答案应该是爆int了,平常不怎么用print…

Computer Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4440    Accepted Submission(s): 2236 Problem Description A school bought the first computer some time ago(so this computer's id is 1). Du…

题目大意:有两排城市,这两排城市之间有一些路相互连接着,求有多少条路相互交叉. 思路:把全部的路先依照x值从小到大排序,x值同样的依照y值从小到大排序,然后插入边的时候,先找有多少比自己y值小的,这些边的x值一定比自己大,也就是一个逆序对,然后统计起来.记得答案要用long long (__int64) CODE: #include <cstdio> #include <cstring> #include <iostream> #include <algorith…

大致题意:在某个点派出两个点去遍历全部的边,花费为边的权值,求最少的花费 思路:这题关键好在这个模型和最长路模型之间的转换.能够转换得到,全部边遍历了两遍的总花费减去最长路的花费就是本题的答案,要思考.并且答案和派出时的起点无关 求最长路两遍dfs或bfs就可以,从随意点bfs一遍找到最长路的一个终点,再从这个终点bfs找到起点 //1032K 79MS C++ 1455B #include<cstdio> #include<iostream> #include<cstrin…