题目链接 题意 给一个\(2\)x\(n\)的矩阵,每个格子看成一个点,每个格子与相邻的格子间有边.现进行一些加边与删边操作,问每次操作后图中有多少条割边. 思路 参考 https://www.cnblogs.com/rpSebastian/p/7834027.html 割边 在这个图中什么样的边才会是割边?情况貌似有点多. 那么满足什么条件的边不会是割边?在环里面的边. 环的要求是什么?第一排和第二排对应位置都有边. 环里面哪些边不是割边?找到这个环中最左边和最右边的竖边,夹起来的一整块里面所…

Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2496    Accepted Submission(s): 788 Problem Description Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack…

Coder Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4838    Accepted Submission(s): 1853 Problem Description In mathematics and computer science, an algorithm describes a set of procedures…

Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030    Accepted Submission(s): 743 Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy t…

1.HDU 5877  Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au,查询它有多少个祖先av满足av<=k/au. (1)dfs+线段树 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3308 题目很好懂,就是单点更新,然后求区间的最长上升子序列. 线段树区间合并问题,注意合并的条件是a[mid + 1] > a[mid],写的细心点就好了. #include <iostream> #include <cstring> #include <cstdio> using namespace std; ; struct SegTree { int l , r…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2795 题目大意:有一块h*w的矩形广告板,要往上面贴广告;   然后给n个1*wi的广告,要求把广告贴上去: 而且要求广告要尽量往上贴并且尽量靠左;  求每个广告的所在的位置,不能贴则为-1. 用线段树模拟,要是左子树的最大值比当前广告大,就查询更新左子树,否则就右子树. #include <iostream> #include <cstdio> #include <cstrin…

Conturbatio Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5480 Description There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place. There are also many quer…

http://acm.hdu.edu.cn/showproblem.php?pid=1828 Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2135    Accepted Submission(s): 1134 Problem Description A number of rectangular posters,…

题意: 给出一个有n个数的数列,并定义mex(l, r)表示数列中第l个元素到第r个元素中第一个没有出现的最小非负整数. 求出这个数列中所有mex的值. 思路: 可以看出对于一个数列,mex(r, r~l)是一个递增序列 mex(0, 0~n-1)是很好求的,只需要遍历找出第一个没有出现的最小非负整数就好了.这里有一个小技巧: tmp = ; ; i <= n; ++i) { mp[arr[i]] = ; while (mp.find(tmp) != mp.end()) tmp++; mex[i…