题目链接: http://acm.hust.edu.cn/vjudge/problem/48416 Shopping Malls Time Limit: 3000MS 问题描述 We want to create a smartphone application to help visitors of a shopping mall and you have to calculate the shortest path between pairs of locations in the mall…

题目链接: POJ:id=3831" target="_blank">http://poj.org/problem?id=3831 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=3264 Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the…

http://acm.hdu.edu.cn/showproblem.php?pid=3264 Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2139    Accepted Submission(s): 775 Problem Description The city of M is a…

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 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…

题目传送门 /* 题意:有n个商店排成一条直线,有一些商店有先后顺序,问从0出发走到n+1最少的步数 贪心:对于区间被覆盖的点只进行一次计算,还有那些要往回走的区间步数*2,再加上原来最少要走n+1步就是答案了 详细解释:http://blog.csdn.net/u013625492/article/details/45640735 */ #include <cstdio> #include <algorithm> #include <cstring> #include…

Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2458    Accepted Submission(s): 906 Problem Description The city of M is a famous shopping city and its open-air shopping…

题意: 给出一些关系用aX <= bY表示, 最后查询aX 和 bY的关系,是>=,==,<=,还是不能确定,还是出现了矛盾. 解法:对每一个关系其实都可以建一条X->Y的边,边权为b/a,表示X <= b/a*Y,输入完后,跑一遍floyd,对每一对点A,B跑出A<=xB的x的最小值. 因为如果真的存b/a的话,可能会损失很大的精度,但是好像这样也能过.为了更加保险,我们可以取个对数,log(b/a) = log(b)-log(a),那么乘法就变成了加法,更好计算,也…

链接 枚举伞的圆心,最多只有20个,因为必须与某个现有的圆心重合. 然后再二分半径就可以了. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set&g…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3264 题意:给你n个圆,坐标和半径,然后要在这n个圆的圆心画一个大圆,大圆与这n个圆相交的面积必须大于等于每个圆面积的一半,问你建在那个圆心半径最小,为多少. 题解:枚举这n个圆,求每个圆的最小半径,通过二分半径来求,然后取这n个的最小值即可,注意点精度就OK了. AC代码: #include <iostream> #include <cstdio> #include <cstri…