Disharmony Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 663    Accepted Submission(s): 307 Problem Description One day Sophia finds a very big square. There are n trees in the square. T…

题意:给你n棵树,每棵树上有两个权值X H 对于X离散化 :3 7 1 5 3 6 -> 2 6 1 4 2 5,对于H一样 然后F = abs(X1-X2)   S=min(H1,H2) 求出每一对F*S的总和 可以看到一边是求每个数与其他数的最小值,一边是求每个数与其他数的差距.因此我们可以排序一边,处理另一边. 我们排序H,因为这样对于固定一个Xi Hi,从小到大每次都是Hi去乘以Xi与剩下的所有X的差的总和. 这样我们就可以使用树状数组维护两个值:每个位置值的个数,每个位置值的总大小,接…

题解:在路边有一行树,给出它们的坐标和高度,先按X坐标排序.记录排名,记为rankx,再按它们的高度排序,记录排名,记为rankh.两颗树i,j的差异度为 fabs(rankx[i]-rankx[j])*min(rankh[i],rankh[j]) 最后求出任异两颗树差异度的和. 题解:首先,需要解决的是min(rh)的问题,对于这个问题,只要离散化之后按rh从大到小排序顺序求解即可,然后用树状数组维护之前出现rx比当前值小的个数与总和,那么同时就可以计算rx比当前大的个数和总和了,那么就可以依…

题意:给出n棵树,给出横坐标x,还有它们的高度h,先按照横坐标排序,则它们的横坐标记为xx, 再按照它们的高度排序,记为hh 两颗树的差异度为 abs(xx[i] - xx[j]) * min(hh[i],hh[j]),求所有的差异度的和 和上面一道题一样,只不过这题是要Min的值,就将h从大到小排序,保证每一个h都是当前最小的 然后维护比当前x小的坐标的个数,当前区间的总和,前面比x小的坐标的和 #include<iostream> #include<cstdio> #inclu…

Disharmony Trees HDU - 3015 One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them. She finds that trees maybe disharmony and the Disharmony Value between two trees is associa…

Disharmony Trees Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.…

/* 写完这篇博客有很多感慨,过去一段时间都是看完题解刷题,刷题,看会题解,没有了大一那个时候什么都不会的时候刷题的感觉,这个题做了一天半,从开始到结束都是从头开始自己构思的很有感觉,找回到当初的感觉 */ Disharmony Trees Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 90 Accepted Submission(s):…

#include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; struct node { ll x,h; } Tr[]; ],tr2[]; bool cmp(node s1,node s2) { return s1.x<s2.x; } bool cmp1(node s1,node s2) { return s1.h<s2.h;…

H - Visible Trees Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2841 Description There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) poi…

Problem Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them. She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with…

Problem Description One day Sophia finds a very big square. There are n trees in the square. They are all so tall. Sophia is very interesting in them.She finds that trees maybe disharmony and the Disharmony Value between two trees is associated with…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2841 题意:给n*m的矩阵(从(1,1)开始编号)格子,每个格子有一棵树,人站在(0,0)的位置,求可以看到多少棵树.同一直线上的树只能看到最靠近人的那颗. 思路:可以将题目转化为求gcd(x, y) = 1,(1 <= x <= n, 1 <= y <= m)的对数.直接套用莫比乌斯反演即可. code: #include <cstdio> #include <cs…

http://www.cnblogs.com/keam37/p/3639294.html keam所有 转载请注明出处 Problem Description Give you two definitions tree and rooted tree. An undirected connected graph without cycles is called a tree. A tree is called rooted if it has a distinguished vertex r c…

标题效果:给你个m*n方格,广场格从(1,1)开始. 在树中的每个点,然后让你(0,0)点往下看,问:你能看到几棵树. 解题思路:假设你的视线被后面的树和挡住的话以后在这条线上的树你是都看不见的啊.挡住的话就是这个小的方格内对角线的连线过顶点,如图: watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveHUxMjExMDUwMTEyNw==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/g…

Visible Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how ma…

Visible Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1951    Accepted Submission(s): 792 Problem Description There are many trees forming a m * n grid, the grid starts from (1,1). Farm…

原文链接 There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see. If two trees and Sherlock are in one line, Farmer Sherlock can only see the tree nearest to…

题目大意:求有n个节点的树有几种? 题解:http://www.cnblogs.com/keam37/p/3639294.html #include <iostream> typedef long long LL; using namespace std; LL f[41]; int cnt[41],n; LL C(LL n,LL m){ m=m<(n-m)?m:(n-m); LL ans=1; for(int i=1;i<=m;i++)ans=ans*(n-i+1)/i; ret…

增量法的最小包围圈算法,不会…… #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> using namespace std; const double EPS = 1e-10; inline int sgn(double x) { return (x > EPS) - (x < -EPS);} s…

/** 大意: 求[1,m], [1,n] 之间有多少个数互素...做了 1695 ,,这题就so easy 了 **/ #include <iostream> #include <cmath> #include <algorithm> using namespace std; ; long long phi[maxn]; long long priD[maxn]; int len ; void euler(long long n){ long long m = (in…

这个题给你一堆树,每棵树的位置x和高度h都给你 f[i]代表这棵树的位置排名,s[i]代表这棵树的高度排名 问你任意两棵树的(f[i] - f[j])*min(s[i],s[j])和 (f[i]-f[i-1])*min(s[i],s[i-1]) + (f[i]-f[i-2])*min(s[i],s[i-2]) 首先看看暴力,肯定过不去,毕竟n2复杂度 看这个式子,其实比s[i]大的那么多项都能合并 合并成 ((f[i] - f[i-1])+ (f[i] - f[i-2]) +.....)*s[i…

题意:有一块(1,1)到(m,n)的地,从(0,0)看能看到几块(如果两块地到看的地方三点一线,后面的地都看不到). 思路:一开始是想不到容斥...后来发现被遮住的地都有一个特点,若(a,b)有gcd(a,b)!= 1,那么就会被遮住.因为斜率k一样,后面的点会被遮住,如果有gcd,那么除一下就会变成gcd = 1的那个点的斜率了.所以问题转化为求gcd不为1有几个点,固定一个点,然后容斥. #include<set> #include<map> #include<queue…

题目链接:戳我 凸包模板 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define MAXN 100010 using namespace std; int n,top; double ans; struct Node{int x,y;}t[MAXN],s[MAXN]; inline bool cmp(stru…

题意: m*n(1<=m,n<=100000)的森林里,起始点在(1,1),某人从(0,0)点开始看,问能看到多少棵树. 题解: 求出1~x中的每个数与1~y的数中互质的数的总和.用素数筛筛出1e5以内的素数.在用这些素数筛出1e5以内每个数的素数因子.最后通过容斥算出与每个数互质的个数. #include <iostream> #include <cstdio> #include <algorithm> #include <vector> #i…

卡特兰数又称卡塔兰数,英文名Catalan number,是组合数学中一个常出现在各种计数问题中出现的数列. 以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)的名字来命名,其前几项为(从第零项开始) : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 656412042…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

http://acm.hdu.edu.cn/showproblem.php?pid=1693 题意:n×m的棋盘求简单回路(可以多条)覆盖整个棋盘的方案,障碍格不许摆放.(n,m<=11) #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; struct H { static const int M=1000007;…

Problem DescriptionMost of us know that in the game called DotA(Defense of the Ancient), Pudge is a strong hero in the first period of the game. When the game goes to end however, Pudge is not a strong hero any more.So Pudge’s teammates give him a ne…

称号: Maple trees Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 177 Accepted Submission(s): 63   Problem Description There are a lot of trees in HDU. Kiki want to surround all the trees with the m…