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, L uni…

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…

题目传送门 /* 二分:搜索距离,判断时距离小于d的石头拿掉 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; ll a[MAXN]; int n, m; bool check(ll d) { ; ; ; i&…

E - River Hopscotch POJ - 3258 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…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11031   Accepted: 4737 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: 5193 Accepted: 2260 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: 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…

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 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…

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…

River Hopscotch http://poj.org/problem?id=3258 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21165   Accepted: 8791 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping…

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…

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…

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…

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: 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.…

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<…

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.…

1650: [Usaco2006 Dec]River Hopscotch 跳石子 Time Limit: 5 Sec  Memory Limit: 64 MB 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 takes…

Description Every year the cows hold an ≤ L ≤ ,,,). Along the river between the starting and ending rocks, N ( ≤ N ≤ ,) more rocks appear, each at an integral distance Di < Di < L). To play the game, each cow in turn starts at the starting rock and…

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 takes place on a long, straight river with a rock at the start and another rock at th…

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, L uni…

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 takes place on a long, straight river with a rock at the start and another rock at th…

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: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…

题目真是不好读,大意例如以下(知道题意就非常好解了) 大致题意: 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 问如今要移除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. //Memory Time //420K 391MS #include<iostream> #include<algorithm> usi…

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,&…

[题意] 牛要到河对岸,在与河岸垂直的一条线上,河中有N块石头,给定河岸宽度L,以及每一块石头离牛所在河岸的距离, 现在去掉M块石头,要求去掉M块石头后,剩下的石头之间以及石头与河岸的最小距离的最大值. [解法] 用二分做,但是开始写了三个版本的二分,全都wa. 无赖看了别人的二分,还是不理解,为什么他们写的就能过. 反复思索后,终于明白了:关键在于题目求的是什么. 做题思想:二分所求的最小距离的最大值mid,记录可以去掉的石头块数cnt(注意:当相邻的石头的距离小于等于mid,就可以去掉),…

嗯... 题目链接:http://poj.org/problem?id=3258 一道很典型的二分答案的题目,和跳石头太像了!! 这道题的题目很显然,求最小中的最大值,注意这道题石头的位置不是从小到大输出的,所以要排序一遍... cnt记录可以跳过的石头个数:检查答案时,如果当前石头与前一个石头之间的距离小于mid,那么直接可以跳过,所以cnt++.如果跳不过去,则更新前一块石头的位置. 如果移走的石头数目小于等于m,说明跳的距离必须或者还可能更大,所以l = mid + 1:否则则要r = m…

 去掉石头 题目大意:一群牛在河上的石头上跳来跳去,现在问你如何通过去掉M个石头,使得牛跳过石头的最短距离变得最大? 这一题比较经典,分治法的经典,二分法可以很方便处理这个问题,我们只要明白比较函数这个东西就可以了. 模板: while (……) { mid = (lb + rb) / ; if (Judge_C(……)) else rb = mid; } while判断条件可以根据是整形还是浮点型灵活变换,Judge_C就是比较函数,几乎所有的分治算法都可以这样归纳,我们只要找到合适的比较函数…