题目链接:http://poj.org/problem?id=3258 River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15753   Accepted: 6649 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully ju…

传送门 https://www.cnblogs.com/violet-acmer/p/9793209.html 题意: 有 N 块岩石,从中去掉任意 M 块后,求相邻两块岩石最小距离最大是多少? 题解: 二分答案(假设答案为res) 定义 l = 0 , r = L ; mid = (l+r)/2 ; 判断当前答案 mid 至少需要去除多少块岩石,如果去除的岩石个数 > M,说明当前答案mid > res,r=mid;反之,说明当前答案 mid <= res , l =mid; AC代码…

描述 http://poj.org/problem?id=2456 有n个小屋,线性排列在不同位置,m头牛,每头牛占据一个小屋,求最近的两头牛之间距离的最大值. Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10095   Accepted: 4997 Description Farmer John has built a new long barn, with N (2 <= N <=…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15273   Accepted: 6465 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9923   Accepted: 4252 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6697   Accepted: 2893 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9263 Accepted: 3994 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The e…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13598   Accepted: 5791 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river.…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9326   Accepted: 4016 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

描述 http://poj.org/problem?id=3258 给出起点和终点之间的距离L,中间有n个石子,给出第i个石子与起点之间的距离d[i],现在要去掉m个石子(不包括起终点),求距离最近的两个石子(包括起终点)之间距离的最大值. River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10841   Accepted: 4654 Description Every year the co…

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 21939 Accepted: 9081 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The…

题目链接:http://poj.org/problem?id=3258 题意:给n个石头,起点和终点也是两个石头,去掉这石头中的m个,使得石头间距的最小值最大. 思路:二分石头间的最短距离,每次贪心地check一下是否满足条件即可,具体看代码. AC代码: #include<iostream> #include<stack> #include<vector> #include<algorithm> #include<cmath> using na…

https://vjudge.net/problem/POJ-3258 二分最小值,判断需要删去的点的个数,如果大于给定,则直接return 0,则说明该数需要再小. 最后注意,起点是0终点是l,起点可以不加进数组,终点必须加进去!! #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath…

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, Lunit…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5473   Accepted: 2379 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement ta…

/** 大意:给定n个点,删除其中的m个点,其中两点之间距离最小的最大值 思路: 二分最小值的最大值---〉t,若有距离小于t,则可以将前面的节点删除:若节点大于t,则继续往下查看 若删除的节点大于m,说明t,过于大,需要减小:若删除的节点小于m说明t过于小了,t需要增大 **/ #include <iostream> #include <algorithm> using namespace std; ]; int main() { long long l,n,m; cin>…

题目:http://poj.org/problem?id=3258 题意: 一条河长度为 L,河的起点(Start)和终点(End)分别有2块石头,S到E的距离就是L. 河中有n块石头,每块石头到S都有唯一的距离 问现在要移除m块石头(S和E除外),每次移除的是与当前最短距离相关联的石头, 要求移除m块石头后,使得那时的最短距离尽可能大,输出那个最短距离. 和3273差不多... #include <iostream> #include <cstdio> #include <…

题目并不难,就是比赛的时候没敢去二分,也算是一个告诫,应该敢于思考…… #include<stdio.h> #include<iostream> using namespace std; int main() { long long n; scanf("%I64d",&n); ,right=1e18,mid,num,m,s; ; while(left<=right) { mid=(left+right)>>; num = ; ;i &l…

https://vjudge.net/problem/POJ-3104 一开始思路不对,一直在想怎么贪心,或者套优先队列.. 其实是用二分法.感觉二分法求最值很常用啊,稍微有点思路的二分就是先推出公式: 对每件衣服:mid = x1(烘干时间)+x2(晾干时间):a[i] <= k*x1+x2:将1式带入2式得 x1>=(a[i]-mid)/(k-1)即每件衣服最少用时位x1向上取整. 注意这里k-1为分母,需要单独考虑k=1的情况 #include<iostream> #incl…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18099   Accepted: 8619 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,..…

地址 别人的代码,自己边界总是控制不好,还不知道哪里错了!思维!这种问题代码越简洁反而越不容易错吧.. #include<stdio.h> #include<algorithm> typedef long long ll; using namespace std; ll n,m,L,a[]; bool bi(ll x){ ll i,cnt=,now=; ;i<=n;i++){ if(a[i]-a[now]<=x)cnt++; else now=i; } ; return…

题目:http://poj.org/problem?id=3258 又A一道,睡觉去了.. #include <stdio.h> #include <algorithm> ]; int s, n, m; bool judge(int mid) { , cnt = ; ; i <= n+; i++) { sum += d[i] - d[i-]; if(sum < mid) cnt++; else sum = ; } if(cnt > m) ; ; } int mai…

https://vjudge.net/problem/POJ-3045 读题后提取到一点:例如对最底层的牛来说,它的崩溃风险=所有牛的重量-(底层牛的w+s),则w+s越大,越在底层. 注意范围lb=-INF. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #includ…

P2855 [USACO06DEC]河跳房子River Hopscotch 二分+贪心 每次二分最小长度,蓝后检查需要去掉的石子数是否超过限制. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 50010 int n,m,L,a[N]; bool check(int lim){ ; ,j=;i<…

POJ3285 River Hopscotch 此题是大白P142页(即POJ2456)的一个变形题,典型的最大化最小值问题. C(x)表示要求的最小距离为X时,此时需要删除的石子.二分枚举X,直到找到最大的X,由于c(x)=m时满足题意,所以最后输出的是ub-1或者lb(lb==ub-1 注意相邻距离小于x的要删除(此处不是小于等于),对于相邻的距离小于x的两个石子,当删除其中一个后,又会产生其他的相邻的石子,直接计数不好计数,不妨用两个标记last,cur,其中last表示上一个石子,cur…

一个不错的二分,注释在代码里 #include <stdio.h> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; ///二分搜索答案,最大化最小值 int main() { int L,n,m; ]; while(~scanf("%d %d %d",&L,&n,&…

poj 2456 Aggressive cows && nyoj 疯牛 最大化最小值 二分 题目链接: nyoj : http://acm.nyist.net/JudgeOnline/problem.php?pid=586 poj : http://poj.org/problem?id=2456 思路: 二分答案,从前到后依次排放m头牛的位置,检查是否可行 代码: #include <iostream> #include <algorithm> #include &…

题目 这道题做了几个小时了都没有做出来,首先是题意搞了半天都没有弄懂,难道真的是因为我不打游戏所以连题都读不懂了? 反正今天是弄不懂了,过几天再来看看... 题意:一个人从1点出发到T点去打boss,这个人有两个属性值,防御值和战斗值,这两个值成反比,为了打赢boss我们要使战斗值最大,于是乎防御值就要最低,但是也不能太低,于是乎这个界限在哪,这就是我们要求的.每条路上都有一个索敌值,防御值必须>=索敌值 才能通过.从1点到T点有很多条通路,我们要找的是:这每一条通路中索敌值最大的中索敌值最小的…

Polycarp has a lot of work to do. Recently he has learned a new time management rule: "if a task takes five minutes or less, do it immediately". Polycarp likes the new rule, however he is not sure that five minutes is the optimal value. He suppo…

River Hopscotch 直接中文 Descriptions 每年奶牛们都要举办各种特殊版本的跳房子比赛,包括在河里从一块岩石跳到另一块岩石.这项激动人心的活动在一条长长的笔直河道中进行,在起点和距离起点 L 远的终点各有一块岩石 (1 ≤ L ≤ 10^9).在起点和终点之间,有 N 块岩石 (0 ≤ N ≤ 50000),每块岩石与起点的距离分别为 Di (0 < Di < L). 在比赛过程中,奶牛轮流从起点出发,尝试到达终点,每一步只能从一块岩石跳到另一块岩石.当然,实力不济的奶…