Rebuild Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 521    Accepted Submission(s): 125 Problem Description Archaeologists find ruins of Ancient ACM Civilization, and they want to rebuild i…

Rebuild Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 446    Accepted Submission(s): 113 Problem Description Archaeologists find ruins of Ancient ACM Civilization, and they want to rebuild i…

题意:n个顶点组成的多边形能否形成正多边形? #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <iostream> #include <algorithm> #include <queue> #include <map> #include <vector> using namespac…

2015 ACM/ICPC 长春现场赛 E题 三分. 如果节点个数是奇数,那么直接列方程可以求解,因为,如果第一个圆半径变大,必然导致最后一个圆的半径变大, 所以,节点是奇数的时候,要么无解,要么只有一组解. 如果节点个数是偶数,如果奇数编号起点的线段长度之和不等于偶数编号起点的线段长度之和,那么必定无解,这个列方程也可以发现的. 剩下的就要三分求最优解了,边界的确定有点小坑,第一个半径是X,第二个半径就是 L12-X,第三个半径就是L23-L12+X.....写出n个关于X的表达式,可以确定三…

题意:给n个点,以这n个点为圆心画圆,使得所有的圆与其相邻的圆相切. 求n个圆最小的面积和. 分析:很容易想到确定了其中一个圆的半径之后,其他的圆的半径也能随之确定了. 画一画三个点的和四个点的,会发现有区别. 三个点的你会发现你稍微画次一点就不能满足 与相邻的都相切的条件了 而四个点的,很轻易就能画出来 所以可以分成两类:奇数个点的 和 偶数个点的 奇数个点的因为答案唯一,因此直接$\frac{R}{2}$就行了 至于偶数个的 因为圆的半径和面积是成单峰函数的.因此可以对半径三分来求解. 其中…

题目大意: 给定一个n边形的顶点 以每个顶点为圆心画圆(半径可为0) 每个顶点的圆要和它相邻顶点的圆相切(不相邻的可相交) 求所有圆的最小面积总和并给出所有圆的半径 设半径为r1 r2 ... rn,顶点距离为L1 L2 ... Ln 当顶点数为奇数时 由 r1+r2=L1 r2+r3=L2 ...... rn+r1=Ln 可得 r1+r1=L1-L2+L3-......+Ln 只要中间所有半径 r >= 0 那么r1就有唯一解 当顶点数为偶数时 由 r1+r2=L1 r2+r3=L2 ....…

The plan of city rebuild Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 616    Accepted Submission(s): 215 Problem Description News comes!~City W will be rebuilt with the expectation to become…

题意  一个城市原来有l个村庄 e1条道路  又添加了n个村庄 e2条道路  后来后销毁了m个村庄  与m相连的道路也销毁了  求使全部未销毁村庄相互连通最小花费  不能连通输出what a pity! 还是非常裸的最小生成树  把销毁掉的标记下  然后prim咯  结果是无穷大就是不能连通的 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N =…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5531 Problem Description Archaeologists find ruins of Ancient ACM Civilization, and they want to rebuild it. The ruins form a closed path on an x-y plane, which has n endpoints. The endpoints locate on (…

HDU 5723 Abandoned country(落后渣国) Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the r…

Abandoned country 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5723 Description An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) ro…

Meeting Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5512 Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two…

City Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 483    Accepted Submission(s): 203 Problem Description After many years, the buildings in HDU has become very old. It need to rebuil…

意甲冠军:要在N好M行和列以及列的数字矩阵和,每个元件的尺寸不超过9,询问是否有这样的矩阵,是独一无二的N(1 ≤ N ≤ 500) , M(1 ≤ M ≤ 500). 主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=4975 -->>方法如:http://blog.csdn.net/scnu_jiechao/article/details/40658221 先做hdu - 4888,再来做此题的时候,感觉这题好 SB 呀.将代码提交后自己就 S…

Abandoned country Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4477    Accepted Submission(s): 1124 Problem Description An abandoned country has n(n≤100000) villages which are numbered from 1…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=3949 题意: 给出n个数,从中任意取几个数字异或,求第k小的异或和 思路: 线性基求第k小异或和,因为题目中可以出现异或和为0的情况,但线性基里是不会出现异或和为0的情况,所以我们需要多处理下,将数字全插入到线性基中,如果无法插入也就代表会出现异或和为0的情况,那么求第k小就应该变成求线性基中第k-1小. 实现代码: #include<bits/stdc++.h> using namespace s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3861 题意:输入t,表示t个样例.接下来每个样例第一行有两个数n,m表示点数和有向边的数量,接下来输入m条有向边,现在要我们把点划分成最少的块,每个块里面的点两两之间要至少有一条有向边可以从其中一个点到另一个点. 思路:分成的区域里面两个点之间至少要有一个点可以到达另一个点,并且要区域数最少,那么就是求最小路径覆盖,但是要求最小路径覆盖的前提是要是无环,所以我们要先用tarjan算法缩点,然后在在缩…

Abandoned country 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5723 Description An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) ro…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5512 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Descriptionn pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mou…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5463 Clarke and minecraft Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 366    Accepted Submission(s): 193 Problem Description Clarke is a patien…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=5723 Abandoned country Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 7573    Accepted Submission(s): 1850 Problem Description An abandoned country…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4348 一开始把lazy标记给push_down了,后来发现这样会让持久化变乱,然后想到不用push_down也可以统计和,改写之后就过了. #include<bits/stdc++.h> using namespace std; ; int a[maxn]; int rt[maxn]; int ts; int n; struct Node { int lson,rson; long long va…

There is a war Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 2435 64-bit integer IO format: %I64d      Java class name: Main There is a sea.There are N islands in the sea.There are some directional bridges…

City Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 503    Accepted Submission(s): 213 Problem Description After many years, the buildings in HDU has become very old. It need to rebui…

http://acm.hdu.edu.cn/showproblem.php?pid=5723 Abandoned country Problem Description   An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=2871 题意: 四种操作: 1.Reset  清空所有内存2.New x  分配一个大小为x的内存块返回,返回能分配的最小的起始点 3.Free x  释放当前点所在的内存块,并输出左右端点 4.Get x  返回第x个内存块的起始点 讨论每个操作的写法: 第一个操作,把线段树初始化就好了 第二个操作,区间合并的基础操作, 第三个操作:多维护两个数组:st,ed代表当前点所属内存块的左右区间 第四个操作:…

Tunnel Warfare HDU 1540 区间合并+最大最小值 题意 D x是破坏这个点,Q x是表示查询以x所在的最长的连续的点的个数,R是恢复上一次破坏的点. 题解思路 参考的大佬博客 这里巧妙使用了最大值最小值来进行区间的查找.上一行是大佬的详细题解,真的很妙啊. 代码实现 #include<cstdio> #include<cstring> #include<algorithm> #include<stack> #define ls (rt&l…

因为之前写的程序比较小,编译起来比较快,所以一直都没有太在意 Build 和 Rebuild 之间的区别,后来发现两个还是有很大不同. Build 只针对在上次编译之后更改过的文件进行编译,在项目比较庞大的时候,Build 还是很有优势的. Rebuild 会编译所有文件,一般相当于执行 Clean + Build(清理并生成). 无论 Build 和 Rebuild,都可以分别针对解决方案和项目.当目标是解决方案的时候,那么 Build 或者 Rebuild 的目标就是解决方案中所有的项目.…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

node-gyp在编译前会首先尝试下载node的headers文件,像这样: gyp http GET https://nodejs.org/download/release/v6.8.1/node-v6.8.1-headers.tar.gz 然后就会卡住,急死人啊 解决方法是:加一个nodedir参数,告诉node-gyp,不需要去网上下载node头文件了,像这样: node-gyp rebuild --nodedir ~/work/node-v6.8.1 问题解决.…