POJ 2456_Aggressive cows】的更多相关文章

题意: 给定N个位置,把C头牛分别放入,求相邻两头牛的最大距离. 分析: 即为求两头牛之间最小距离的最大值.二分搜索答案. 代码: #include<iostream> #include<algorithm> #include<cstdio> using namespace std; const int maxn = 100005, INF =0x3fffffff; typedef long long ll; int x[maxn]; int N, C; int jud…
题目传送门 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; int cnt[MAX_N]; int ans[MAX_N]; ; struct node { int s, e; int id; }cow[MAX_N]; inline int read(void) { , f = ; char ch = getchar (); ; ch = getchar…
Cows Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 16546   Accepted: 5531 Description Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in hi…
Cows Time Limit: 3000MS Memory Limit: 65536K Description Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. Farmer John ha…
Cows 题目:http://poj.org/problem?id=2481 题意:有N头牛,每仅仅牛有一个值[S,E],假设对于牛i和牛j来说,它们的值满足以下的条件则证明牛i比牛j强壮:Si <=Sjand Ej <= Ei and Ei - Si > Ej - Sj. 如今已知每一头牛的測验值,要求输出每头牛有几头牛比其强壮. 思路:将牛依照S从小到大排序.S同样依照E从大到小排序,这就保证了排在后面的牛一定不比前面的牛强壮. 再依照E值(离散化后)建立一颗线段树(这里最值仅仅有1…
                                                                  Cows Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 17626   Accepted: 5940 Description Farmer John's cows have discovered that the clover growing along the ridge of the h…
求凸包面积.求结果后不用加绝对值,这是BBS()排序决定的. //Ps 熟练了template <class T>之后用起来真心方便= = //POJ 3348 //凸包面积 //1A 2016-10-15 #include <cstdio> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #define MAXN (10000 +…
题目链接:http://poj.org/problem?id=2186 题目大意:有n头牛和m对关系, 每一对关系有两个数(a, b)代表a牛认为b牛是“受欢迎”的,且这种关系具有传递性, 如果a牛认为b牛“受欢迎”, b牛认为c牛“受欢迎”, 那么a牛也认为c牛“受欢迎”. 现在想知道有多少头牛受除他本身外其他所有牛的欢迎? 解题思路:如果有两头或者多头牛受除他本身外其他所有牛的欢迎, 那么在这两头或者多头牛之中, 任意一头牛也受两头或者多头牛中别的牛的欢迎, 即这两头或者多头牛同属于一个强联…
题目链接:http://poj.org/problem?id=2481 给你n个区间,让你求每个区间被真包含的区间个数有多少,注意是真包含,所以要是两个区间的x y都相同就算0.(类似poj3067,cf652D) 对每个区间的x从小到大排序,相同的话按y从大到小排序.然后对枚举每个区间的y求其逆序对,然后在y的位置上置1.但是存在两个区间完全重合,我的做法比较搓,就是判断和前一个区间是否完全相同,要是相同,就把前一个答案赋值给这个. #include <iostream> #include…
Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9851   Accepted: 3375 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free…
Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9022   Accepted: 3992 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…
题链: http://poj.org/problem?id=3348 题解: 计算几何,凸包,多边形面积 好吧,就是个裸题,没什么可讲的. 代码: #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 10050 using namespace std; const double eps=1e-8…
[题解] 参考https://blog.csdn.net/acmer_hades/article/details/46272605.设置数组pre_smaller,其中第i个元素即为输入的第i项,则显然pre_smaller[1] = 0.build_tree建立树结构,分解区间[1, N],其中每个节的度要么为2,要么为0.query中参数smaller_num表示自身以及队列前方比自己编号小的cows的总数.确实是非常简洁的方法,将线段数组运用得淋漓尽致!不愧是NOI大神 [代码] #inc…
Description Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. Farmer John has N cows (we number the cows from 1 to N). Ea…
Cows Time Limit: 2000MS Memory Limit: 65536K Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts b…
LINK 题意:给出点集,求凸包的面积 思路:主要是求面积的考察,固定一个点顺序枚举两个点叉积求三角形面积和除2即可 /** @Date : 2017-07-19 16:07:11 * @FileName: POJ 3348 凸包面积 叉积.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <…
Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6199   Accepted: 2822 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…
Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7038   Accepted: 3242 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…
Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possibl…
Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possibl…
看的人家的思路,没有理解清楚,,, 结果一直改一直交,,wa了4次才交上,,, 注意: 为了使用树状数组,我们要按照e从大到小排序.但s要从小到大.(我开始的时候错在这里了) 代码如下: #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <cstdlib> #include <stack> #include <que…
题目传送门 题意:求凸包 + (int)求面积 / 50 /************************************************ * Author :Running_Time * Created Time :2015/11/4 星期三 11:13:29 * File Name :POJ_3348.cpp ************************************************/ #include <cstdio> #include <a…
题意:对于两个区间,[si,ei] 和 [sj,ej],若 si <= sj and ei >= ej and ei - si > ej - sj 则说明区间 [si,ei] 比 [sj,ej] 强.对于每个区间,求出比它强的区间的个数. 解题思路:先将每个区间按 e 降序排列,在按 s 升序排列.则对于每个区间而言,比它强的区间的区间一定位于它的前面. 利用数状数组求每个区间[si,ei]前面 满足条件的区间[sj,ej]个数(条件:ej<=ei),再减去前面的和它相同的区间的个…
题目链接 大意: 求凸包的面积. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue>…
题目大意: 给你n棵树,可以用这n棵树围一个圈,然后在圈里面可以养牛,每个牛需要50平方米的空间,问最多可以养多少牛? 其实就是求一个凸包,计算凸包面积,然后除以50,然后就得到答案,直接上模板了. 凸包这一类型的题目差不多,可以作为模板使用,时间复杂度是NlogN. //Time 32ms; Memory 568K #include<iostream> #include<algorithm> using namespace std; int n; typedef struct p…
题意: 给出一个序列a[1....n],a[i]代表在0....i-1中比a[i]小的个数. 求出这个序列. 思路: 1:暴力. #include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #include<queue> #include<deque> #i…
<题目链接> 题目大意: 就是给出N个区间,问这个区间是多少个区间的真子集. 解题分析: 本题与stars类似,只要巧妙的将线段的起点和终点分别看成 二维坐标系中的x,y坐标,就会发现,其实本题就是求每个点(把线段看成点) 左上角点的个数(包括边界,但并不包括与该点坐标完全相同的点),所以,与stars类似,对所有线段先进行排序,按 y坐标由大到小排序,若左边相同,就对x坐标进行从小到大排序.然后就可以直接对每个点的x坐标建立一维树状数组求解了. #include <cstdio>…
Description 作为对奶牛们辛勤工作的回报,Farmer John决定带她们去附近的大城市玩一天.旅行的前夜,奶牛们在兴奋地 讨论如何最好地享受这难得的闲暇. 很幸运地,奶牛们找到了一张详细的城市地图,上面标注了城市中所有 L(2 <= L <= 1000)座标志性建筑物(建筑物按1..L顺次编号),以及连接这些建筑物的P(2 <= P <= 5000)条道路 .按照计划,那天早上Farmer John会开车将奶牛们送到某个她们指定的建筑物旁边,等奶牛们完成她们的整个 旅行…
想当年--还是邱神给我讲的凸包来着-- #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define eps 0.000000001 #define enter putchar('\n') #define space putchar(' ') using namespace std; typedef long long ll; template <c…
题目: 给几个点,用绳子圈出最大的面积养牛,输出最大面积/50 题解: Graham凸包算法的模板题 下面给出做法 1.选出x坐标最小(相同情况y最小)的点作为极点(显然他一定在凸包上) 2.其他点进行极角排序<极角指从坐标轴的某一方向逆时针旋转到向量的角度>, 极角一样按距离从近到远(可以用叉积实现) 3.用栈维护凸包上的点,将极点和极角序最小的点依次入栈 4.按顺序扫描,检查栈顶的前两个元素与这个点构成的线段是否拐向右(顺时针侧,叉积小于0) 如果满足就弹出栈顶元素,直到不满足或者栈里不足…