POJ-3258-River Hopscotch
来源:互联网 发布:python dict索引 编辑:程序博客网 时间:2024/05/29 13:21
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 the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distanceDi from the start (0 < Di < L).
To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.
Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).
FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.
Input
Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.
Output
Sample Input
25 5 2214112117
Sample Output
4
Hint
#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>#include<algorithm>#include<queue>#include<vector>#include<map>using namespace std;int main(){ int l, n, m, high, mid, low, stone[50010], cnt, pre, ans; scanf("%d%d%d", &l, &n,&m); for(int i = 1; i<=n; ++i) scanf("%d", &stone[i]); sort(stone+1, stone+1+n); stone[0] = 0, stone[n+1] = l; high = l, low = 0; while(low<=high) { mid = (high+low)/2; cnt = pre = 0; for(int i = 1; i<=n+1; ++i) { if(stone[i]-stone[pre]<mid) cnt++; else pre = i; } if(cnt>m) high = mid-1; else { low = mid+1; ans = mid;// 存储mid的值, 在下一循环中可能在if语句中跳出,而不会走else, 此时mid值已经改变 } } printf("%d\n", ans); return 0;}
- poj 3258 River Hopscotch
- poj-3258 River Hopscotch
- poj-3258 River Hopscotch
- poj 3258 River Hopscotch
- poj 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- poj 3258 River Hopscotch
- POJ-3258-River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- poj 3258-River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258--River Hopscotch
- poj 3258 River Hopscotch
- 顺序表应用3:元素位置互换之移位算法
- 数据库视图view简介
- 设计模式之禅——代理模式(一)普通代理&强制代理&虚拟代理&动态代理
- Swift 枚举
- (LeetCode)Palindrome Linked List --- 单链表回文
- POJ-3258-River Hopscotch
- android虚线
- 将现有的SQL工作负载迁移至hadoop竟然如此简单!
- Android fragment间传递数据以及Dialog
- C#中&结&构&体&与&字&节&流&互&相&转&换
- windows删除node_modules[文件名或扩展名太长,目录层级太深]
- zabbix监控系统fping配置 提示不支持的
- HTTP协议详解
- Java面向对象