Jumping Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6398   Accepted: 3828 Description Farmer John's cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump. The…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25944   Accepted: 11982 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,.…

Jumping Cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7329 Accepted: 4404 Description Farmer John's cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump. The loc…

POJ 3190 Stall Reservations贪心 Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obvi…

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 &…

POJ 2456 Agressive cows 农夫 John 建造了一座很长的畜栏,它包括N (2≤N≤100,000)个隔间,这 些小隔间的位置为x0,...,xN-1 (0≤xi≤1,000,000,000,均为整数,各不相同). John的C (2≤C≤N)头牛每头分到一个隔间.牛都希望互相离得远点省得 互相打扰.怎样才能使任意两头牛之间的最小距离尽可能的大,这个最 大的最小距离是多少呢 思想:二分,首先把输入的数据进行从小到大排序,再由最短距离为1,最长距离为(a[n-1] - a[0…

POJ 2392 Space Elevator(贪心+多重背包) http://poj.org/problem?id=2392 题意: 题意:给定n种积木.每种积木都有一个高度h[i],一个数量num[i].另一个限制条件,这个积木所在的位置不能高于limit[i],问能叠起的最大高度? 分析: 本题是一道多重背包问题, 只是每一个物品的选择不只要受该种物品的数量num[i]限制, 且该物品还受到limit[i]的限制. 这里有一个贪心的结论: 我们每次背包选取物品时都应该优先放置当前limit…

题目传送门 /* 二分搜索:搜索安排最近牛的距离不小于d */ #include <cstdio> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int x[MAXN]; int n, m; bool check(int d) { ; ; i<=m-; ++i) { ; while (cur <= n && x[…

poj 2186 Popular Cows 题意: 有N头牛, 给出M对关系, 如(1,2)代表1欢迎2, 关系是单向的且能够传递, 即1欢迎2不代表2欢迎1, 可是假设2也欢迎3那么1也欢迎3. 求被全部牛都欢迎的牛的数量. 限制: 1 <= N <= 10000 1 <= M <= 50000 思路: Kosaraju算法, 看缩点后拓扑序的终点有多少头牛, 且要推断是不是全部强连通分量都连向它. Kosaraju算法.分拆完连通分量后,也完毕了拓扑序. /*poj 2186…

缩点练习 洛谷 P3387 [模板]缩点 缩点 解题思路: 都说是模板了...先缩点把有环图转换成DAG 然后拓扑排序即可 #include <bits/stdc++.h> using namespace std; /* freopen("k.in", "r", stdin); freopen("k.out", "w", stdout); */ //clock_t c1 = clock(); //std::cerr…