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…

题目传送门 题意:n头牛,1~n的id给它们乱序编号,已知每头牛前面有多少头牛的编号是比它小的,求原来乱序的编号 分析:从后往前考虑,最后一头牛a[i] = 0,那么它的编号为第a[i] + 1编号:为1,倒数第二头牛的编号为除去最后一头牛的编号后的第a[i-1] + 1编号:为3,其他的类推,所以可以维护之前已经选掉的编号,求第k大的数字,sum[rt] 表示该区间已经被选掉的点的个数.另外树状数组也可以做,只不过用二分优化查找第k大的位置. 收获:逆向思维,求动态第K大 代码(线段树): /…

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers wa…

一道dfs序+树状数组的题 因为并没有get到dfs序以及对树状数组也不熟练卡了很久orz dfs序: in和out是时间戳 dfs序可以将树转化成为一个序列,满足区间 -> 子树 然后就可以用树状数组之类的维护序列的东东来维护了 ; void dfs(int u, int fa) { seq[++idx] = u; in[u] = idx; for(int i = head[u];i;i = nxt[i]) { int v = l[i]; if(v != fa) { dfs(v, u); }…

题目传送门 题意:两种操作,问u到v的距离,并且u走到了v:把第i条边距离改成w 分析:根据DFS访问顺序,将树处理成链状的,那么回边处理成负权值,那么LCA加上BIT能够知道u到v的距离,BIT存储每条边的信息,这样第二种操作也能用BIT快速解决 利用RMQ的写法不知哪里写挫了,改用倍增法 /************************************************ * Author :Running_Time * Created Time :2015/10/6 星期二…

Cows Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 14906   Accepted: 4941 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 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>…