Chook_lxk的博客

点击打开题目链接1491: [NOI2007]社交网络Time Limit:10 SecMemory Limit:64 MBSubmit:1963Solved:1058[Submit][Status][Discuss]Description在社交网络（socialnetwork）的研究中，我们常常使用图论概念去解释一些社会现象。不妨看这样的一个问

点击打开题目链接B. Personal programming languagetime limit per test1.0 smemory limit per test64 MBinputstandard inputoutputstandard outputEverybody knows that you

点击打开题目链接Yukari's BirthdayTime Limit: 12000/6000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6368Accepted Submission(s): 1539Problem Descript

点击打开题目链接Samuel F. B. Morse is best known for the coding scheme that carries his name. Morse code is still usedin international radio communication. The coding of text using Morse code is straightfor

点击打开题目链接Qin Shi Huang's National Road SystemTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8430Accepted Submission(s): 2987

点击打开题目链接题目大意：给出一个N长度的数字序列，对于任一子序列中，都存在唯一不重复出现的数字，则称为不无聊序列，否则为无聊序列。思路：如果某个数字a[i]在序列中只出现一次，那么任何包含a[i]的序列都是不无聊的,则只需要考虑i-1的序列和i+1的序列是否满足。以此类推，递归下去。问题则转变成判断某个数字在某子序列中是否唯一。方法是用l[i],r[i]两个数组分别记录总序列

点击打开题目链接题目大意：有N个节点，初始时没有根节点，每次进行I或者E操作。E u : 算出u到根节点的距离I u v : 将v作为u的根节点，之间距离为 abs(u-v)%1000；思路：因为加权的并查集接触的不多。所有拿到题之后并查集的思想只存在大脑1s，就开是dfs找路径。因为终止条件看做了数字0，提交一直WA，没有超时，以为方法正确（TAT），直接WA到比赛结束。

点击打开题目链接N friends go to the local super market together. The probability of their buying something from themarket is p1, p2, p3, . . . , pN respectively. After their marketing is finished you are gi

点击打开题目链接Hou Yi's secretTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4931Accepted Submission(s): 1120Problem Description

点击打开题目链接Isabella's MessageTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2579Accepted Submission(s): 762Problem Descripti

点击打开题目链接A hard Aoshu ProblemTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 62768/32768 K (Java/Others)Total Submission(s): 1550Accepted Submission(s): 813Problem Descrip

点击打开题目链接The Little Girl who Picks MushroomsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2885Accepted Submission(s): 937

点击打开题目链接题目大意：给出一个N*N的01矩阵，将矩阵中的t个0换成1，使得每一个数的四个方向的数之和为偶数。求最小的t.思路：只枚举第一行，然后推出下几行的数值。枚举第一行用到了位运算的方法.附上AC代码：#include#include#includeusing namespace std;const int INF=1e9;int grid[20][2

点击打开题目链接I'm Telling the TruthTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2448Accepted Submission(s): 1204Problem Descr

点击打开题目链接D2. Magic Powder - 2time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe term of this problem

点击打开题目链接题目大意：给出长为l,宽为w的一块长方形草地。草地上有n个喷头，给出每个喷头的圆心坐标及半径。求解最少需要多少喷头覆盖整个长方形。思路：将圆形区间转化成长方形区间。每个圆与草地相交的四个点为长方形的四个顶点。然后贪心求解。附上AC代码：#include#include#includeusing namespace std;const int max