POJ 1700 cross river (数学模拟)】的更多相关文章

                                                                                                                           Crossing River Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10311   Accepted: 3889 Description A group of N p…
Crossing River Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9585 Accepted: 3622 Description A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangem…
Crossing River Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9887   Accepted: 3737 Description A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arra…
http://poj.org/problem?id=1700 题目大意: 有n个人要过坐船过河,每一个人划船有个时间a[i],每次最多两个人坐一条船过河.且过河时间为两个人中速度慢的,求n个人过河的最短时间. 思路: 贪心. 对于每次过河的,有两种情况: //最快和最慢过去,然后最快回来.在和次慢过去.最快回来 int action1=a[i-1] + a[0] + a[i-2] +a[0]; //最快和次慢过去,然后最快回来,在次慢和最慢过去,次慢回来 int action2=a[1] +a[…
Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13982   Accepted: 5349 Description A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must b…
Description A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Eac…
http://poj.org/problem?id=1700 题目大意就是一条船,有N个人需要过河,求N个人最短过河的时间 #include <stdio.h> int main() { ],T,j,i,n,sum; scanf("%d",&T); ;i<T;i++) { scanf("%d",&n); ;j<n;j++) scanf("%d",&t[j]); sum=; ) { ]+t[]+t[n…
Description A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Eac…
POJ题目链接:http://poj.org/problem?id=1700 N个人过河,船每次最多只能坐两个人,船载每个人过河的所需时间不同,问最快的过河时间. 思路: 当n=1,2,3时所需要的最小时间很容易求得,现在由n>=4,假设n个人单独过河所需要的时间存储在数组t中,将数组t按升序排序,那么 这时将单独过河所需要时间最多的两个旅行者送到对岸去,有两种方式:       1> 最快的(即所用时间t[0])和次快的过河,然后最快的将船划回来,再次慢的和最慢的过河,然后次快的将船划回来.…
题目链接: http://poj.org/problem?id=1700 1. 当1个人时: 直接过河 t[0]. 2. 当2个人时: 时间为较慢的那个 t[1]. 3. 当3个人时: 时间为 t[0]+t[1]+t[2]. 4. 当4个以上的人时, 将t[0] 和 t[1]当作搬运工. 两种方案: - 最快和次快的过去, 最快的回来, 最慢和次慢的过去, 次快的回来, 这样就将最慢和次慢的送过去了. 时间: t[0]+2*t[1]+t[i] - 最快的依次送最慢和次慢的过去再回来, 时间: 2…
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…
题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82974#problem/E 解题思路:当n>=4,假设n个人单独过河所需要的时间存储在数组t中,将数组t按升序排序,那么 这时将单独过河所需要时间最多的两个旅行者送到对岸去,有两种方式:       1方案. 最快的(即所用时间t[0])和次快的过河,然后最快的将船划回来,再次慢的和最慢的过河,然后次快的将船划回来.           即所需时间为:t[0]+t[1]+t[…
[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. Sample Input 25 5 2 2 14 11 21 17 Sample Output 4分析:其实这个和Agressive Cows这道题差不多,只要把起点与终点加入到数组中,然后留下找(N-M+2)个位置之间的最大值,类比与Agressive cows的M:具体题解可看Agressi…
Bridge over a rough river Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4143   Accepted: 1703 Description A group of N travelers (1 ≤ N ≤ 50) has approached an old and shabby bridge and wishes to cross the river as soon as possible. Ho…
小需求是当你在第一个下拉框选择了国家时,会自动更新第二个省份的下拉框,效果如下 两个下拉选择Html如下: <select id="country_select"> <option> All Countries</option> <option> Afghanistan</option> <option> Albania</option> <option> Algeria</optio…
问题描述:3个商人带着3个仆人过河,过河的工具只有一艘小船,只能同时载两个人过河,包括划船的人.在河的任何一边,只要仆人的数量超过商人的数量,仆人就会联合起来将商人杀死并抢夺其财物,问商人应如何设计过河顺序才能让所有人都安全地过到河的另一边. 详细过程参见<数学模型>第四版(姜启源) #include <cstdio> #define maxn 101 int num;//number of bus or fol int graph[maxn*maxn][maxn*maxn]; i…
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…
A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has…
Description One of the more popular activities in San Antonio is to enjoy margaritas in the park along the river know as the River Walk. Margaritas may be purchased at many establishments along the River Walk from fancy hotels to Joe’s Taco and Marga…
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…
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…
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…
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…
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…
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…