最小最大...又是经典的二分答案做法.. -------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #def…

1734: [Usaco2005 feb]Aggressive cows 愤怒的牛 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,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C &l…

二分答案,贪心判定 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N=100005; int n,m,a[N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(p>=…

1734: [Usaco2005 feb]Aggressive cows 愤怒的牛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 217  Solved: 175[Submit][Status][Discuss] Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a st…

[Usaco2005 feb]Aggressive cows 愤怒的牛 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 407  Solved: 325[Submit][Status][Discuss] Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight…

水题,20分钟AC,最大值最小,一看就是二分答案... 代码: Description Farmer John has built a <= N <= ,) stalls. The stalls are located along a straight line at positions x1,...,xN ( <= xi <= ,,,). His C ( <= C <= N) cows don't like this barn layout and become ag…

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,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C <= N) cows don't like this barn layout an…

一道水题WA了这么多次真是.... 统考终于完 ( 挂 ) 了...可以好好写题了... 先floyd跑出各个点的最短路 , 然后二分答案 m , 再建图. 每个 farm 拆成一个 cow 点和一个 shelter 点, 然后对于每个 farm x : S -> cow( x ) = cow( x ) 数量 , shelter( x ) -> T = shelter( x ) 容量 ; 对于每个dist( u , v ) <= m 的 cow( u ) -> shelter( v…

话说二分和三分的题还没有整理过,就趁这两题来整理下笔记 先讲讲关于二分,对于二分的具体边界长期以来对我来说都是个玄学问题,都是边调边拍改对的.思路大体是确定左边界l,和有边界r,判断满足条件缩小范围. 放个大概的代码 while(l+ep<r){ lm=l+(r-l)/3.0; rm=r-(r-l)/3.0; if(clu(lm)>clu(rm)) l=lm; else r=rm; } 二分用处很大,一般用在二分答案以及二分查找,一般看到最大的最小或最小的最大都是二分答案或二分查找题,一般来说…

题目链接:BZOJ - 1733 题目分析 直接二分这个最大边的边权,然后用最大流判断是否可以有 T 的流量. 代码 #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MaxN = 200 + 5,…