POJ 2481 Cows】的更多相关文章

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 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: 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…
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…
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 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…
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…
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> 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>…