River Hopscotch(二分POJ3258)】的更多相关文章

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…
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: 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 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…
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…
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 <…
/** 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 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…