题意:给定正方形的边长 r ,在平面内寻找正方形可以圈住的点的最大的个数. 分析:先对点排序,然后固定一条边,再平移另一条垂直边,得到点的个数,最后比较大小即可. 注意:不包含正方形倾斜的情况! // Time 125ms; Memory 240k #include<iostream> #include<cstdio> #include<algorithm> using namespace std; struct point { int x,y; }; bool ope…

题目链接 网络赛的水实在太深,这场居然没出线zzz,差了一点点,看到这道题的的时候就剩半个小时了.上面是官方的题意题解,打完了才知道暴力就可以过,暴力我们当时是想出来了的,如果稍稍再优化一下估计就过了zzz.去年有一场现场赛也是n=1000,n^3过了,看来关键时刻实在做不出来就得大胆暴力啊. #include <bits/stdc++.h> using namespace std; typedef long long ll; ; int a[maxn],nex[maxn]; int main…

BFS+链表 代码改自某博客 #include<stdio.h> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<map> #include<set> #include<vector> #include<queue> using nam…

HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863    Accepted Submission(s): 342 Problem Description Kaname Mado…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 271    Accepted Submission(s): 121 Problem Description It's time for music! A lot of popular musicians are invited to join us in the…

Poker Shuffle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 95    Accepted Submission(s): 24 Problem Description Jason is not only an ACMer, but also a poker nerd. He is able to do a perfect s…

Divide Groups Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 423    Accepted Submission(s): 161 Problem Description   This year is the 60th anniversary of NJUST, and to make the celebration mor…

Two Rabbits Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 505    Accepted Submission(s): 260 Problem Description Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny…

F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 382    Accepted Submission(s): 137 Problem Description For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x…

Minimum palindrome Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 260    Accepted Submission(s): 127 Problem Description Setting password is very important, especially when you have so many "in…

题目链接  2016 青岛网络赛  Problem C 题意  给出一些敏感词,和一篇文章.现在要屏蔽这篇文章中所有出现过的敏感词,屏蔽掉的用$'*'$表示. 建立$AC$自动机,查询的时候沿着$fail$指针往下走,当匹配成功的时候更新$f[i]$ $f[i]$表示要屏蔽以第$i$个字母结尾的长度为$f[i]$的字符串. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i &l…

签到题都要想一会 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #define MOD 1000000007 const int INF=0x3f3f3f3f; ; typedef long…

题意: 有一幅图,现在要加一条边,加边之后要你删除一条边,使图不连通,费用为边的费用,要你求的是删除的边的最小值的最大值(每次都可以删除一条边,选最小的删除,这些最小中的最大就为答案) 首先要进行缩点,把图缩为一棵树,因此,加入一条边后图就会存在一个环,环中的任何一条边删除后都不会导致图不连通 之后找一条最小的边,可以说这条边肯定是在加边之后的连通块里的,因为如果不在连通块里,那就直接可以把这条最小的边删掉,而达不到求出答案的目的 找到边后,分别从边的两点开始遍历,要遍历出一条路径来,并且边上的…

题意:一个长为L的河,中间有n个石子,小青蛙需要跳少于m次过河,判断小青蛙每次跳跃最大距离的最小值 最大值最小,用二分 Sample Input 6 1 2 2 25 3 3 11 2 18 Sample Output 4 11   2015-07-27:备战区域赛专题 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath>…

题意:求K个机器人从同一点出发,遍历所有点所需的最小花费 链接:点我 Sample Input 3 1 1 //3个点,从1出发,1个机器人 1 2 1 1 3 1 3 1 2 1 2 1 1 3 1 Sample Output 3 2 转移方程: dp[i][j]=min(dp[i][j],dp[i][j*k],dp[son[i]][k]+len(i,son[i])*k) 方程还是比较好写的,主要是要遍历所有的点 下面我们分析一下第一个样例 1 / ＼ / ＼ 2 3 我们派了一个机器人去3,…

题意:求1-n内最大的x/phi(x) 通式:φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数,x是不为0的整数.φ(1)=1(唯一和1互质的数就是1本身). 因此含质因数最多的即为所求,打表求出前n个积,之后找到比自己小的最大积 大数打表 2015-07-27:到此一游 #include<cstdio> #include<iostream> #include<algorith…

题意:给一些指定长宽高的砖,求能累出的最大高度,不同砖有不同编号,每种编号对下面的砖做出了限制 dp 注意输出要用%I64d,否则会wa,以后不用%lld了 Sample Input 3 10 10 12 0 10 10 12 1 10 10 11 2 2 10 10 11 1 10 10 11 1 0Sample Output 24 11 #include<cstdio> #include<iostream> #include<algorithm> #include&…

Problem Description Recently, Dave is boring, so he often walks around. He finds that some places are too crowded, for example, the ground. He couldn't help to think of the disasters happening recently. Crowded place is not safe. He knows there are N…

Dave Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 4095    Accepted Submission(s): 1390 Problem Description Recently, Dave is boring, so he often walks around. He finds that some places are to…

// Time 234 ms; Memory 244 K #include<iostream> #include<cstdio> #include<cmath> using namespace std; typedef struct point { double x,y,z; point(double xx=0,double yy=0,double zz=0):x(xx),y(yy),z(zz){} }vector; vector operator - (point a…

// Time 1718 ms; Memory 1500 K #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define eps 1e-10 #define sqr(a) ((a)*(a)) #define pi (2.0*asin(1.0)) using namespace std; double ma[100010]; int sig(double a)…

#include<stdio.h> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<map> #include<set> #include<vector> #include<queue> #define M(a,b) memset(a,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4737 题目大意:给定一系列数,F(i,j)表示对从ai到aj连续求或运算,(i<=j)求F(i,j)<=m的总数. 解题思路:或运算只会让值变大或保持不变.不断通过右移j来更新F(i,j),当aj>=m时所有的i<=j F(i,j)都大于等于m,因此从j后面继续扫数组:当aj<m而F(i,j)>=m时通过右移i来使F(i,j)<m.扫完整个数组即可. #include…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6197 题意:给你n个数,问让你从中删掉k个数后(k<=n),是否能使剩下的序列为非递减或者非递增序列 解法:签到题,就是让你求最长不下降子序列长度len,然后判断下n-len是否小于k(将序列反着存下来然后再求即最长不上升子序列,取两者len中的较大值),然后直接套nlogn的模板即可. #include <bits/stdc++.h> using namespace std; const…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6194 题意:告诉你一个字符串和k , 求这个字符串中有多少不同的子串恰好出现了k 次. 解法:后缀数组.我们先考虑至少出现k 次的子串, 所以我们枚举排好序的后缀i (sa[i]) .然后k段k 段的枚举.假设当前枚举的是 sa[i]~sa[i + k -1],那么假设这一段的最长公共前缀  是L 的话.那么就有L 个不同的子串至少出现了k次.我们要减去至少出现k + 1次的 , 但还要和这个k 段…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5443 The Water Problem Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1084    Accepted Submission(s): 863 Problem Description In Land waterless…

the Sum of Cube Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 405    Accepted Submission(s): 224 Problem Description A range is given, the begin and the end are both integers. You should sum…

HDU 5045 5047 5050 5053 太菜了,名额差点没保住.吓尿..赶紧开刷树链抛分 5045:状压DP.压缩10个人.因为两个人不能差2以上,所以能够用01表示 5047:推推公式就可以,每次交线多4条 5050:求GCD.用java大叔就可以 5053:签到题 代码: 5045: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int…

题目链接 #include <iostream> #include <math.h> #include <stdio.h> #include<algorithm> #include<cstring> #include<queue> using namespace std; ]; int main() { int T,n,t; scanf("%d",&T); while(T--) { int tmp; sca…

题目链接 题意:有一壶水, 体积在 L和 R之间, 有两个杯子, 你要把水倒到两个杯子里面, 使得杯子水体积几乎相同(体积的差值小于等于1), 并且使得壶里剩下水体积不大于1. 你无法测量壶里剩下水的体积, 问最小需要倒水的次数. 题解:考虑倒水的大致过程,L = 0 和 L = 1 的情况应该是等价的,所以不妨设 L > 0.首先向一个杯子倒 L/2 升水,再往另一个杯子倒  L/2+1 升水.接下来就来回往两个杯子里倒 2 升,直到倒空为止.这样就很容易分析出需要倒水的次数.唯一注意的是最后…