先预处理出来每个点对之间的最短距离 然后二分答案,网络流判断是否可行就好了恩 /************************************************************** Problem: 1738 User: rausen Language: C++ Result: Accepted Time:404 ms Memory:9788 kb **************************************************************…

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…

题目描述 FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a…

Description 约翰的牛们非常害怕淋雨,那会使他们瑟瑟发抖．他们打算安装一个下雨报警器,并且安排了一个撤退计划．他们需要计算最少的让所有牛进入雨棚的时间．    牛们在农场的F(1≤F≤200)个田地上吃草．有P(1≤P≤1500)条双向路连接着这些田地．路很宽,无限量的牛可以通过．田地上有雨棚,雨棚有一定的容量,牛们可以瞬间从这块田地进入这块田地上的雨棚    请计算最少的时间,让每只牛都进入雨棚． Input 第1行:两个整数F和P; 第2到F+1行:第i+l行有两个整数描述第i个田…

Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to cr…

题目要求所有牛都去避雨的最长时间最小. 显然需要二分 二分之后考虑如何判定. 显然每头牛都可以去某个地方 但是前提是最短路径<=mid. 依靠二分出来的东西建图.可以发现这是一个匹配问题 dinic即可. 注意数组要开两倍 因为要拆点 我开小了wa了两发才意识过来. const int MAXN=410; int n,m,S,len,T,L,R,ans; ll a[MAXN][MAXN]; int c[MAXN],cc[MAXN],cur[MAXN],vis[MAXN],q[MAXN]; int…

Ombrophobic Bovines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18623   Accepted: 4057 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They h…

Ombrophobic Bovines Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 14519Accepted: 3170 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have de…

题目 1739: [Usaco2005 mar]Space Elevator 太空电梯 Time Limit: 5 Sec  Memory Limit: 64 MB Description The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) differe…

1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 119  Solved: 100[Submit][Status][Discuss] Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10…

1682: [Usaco2005 Mar]Out of Hay 干草危机 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 391  Solved: 258[Submit][Status] Description The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms…

Ombrophobic Bovines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19359   Accepted: 4186 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They h…

Ombrophobic Bovines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11651   Accepted: 2586 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They h…

Ombrophobic Bovines Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 21660Accepted: 4658 题目链接:http://poj.org/problem?id=2391 Description: FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes the…

Ombrophobic Bovines Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16539 Accepted: 3605 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have…

                     Ombrophobic Bovines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18205   Accepted: 3960 Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in…

1680: [Usaco2005 Mar]Yogurt factory Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 106  Solved: 74[Submit][Status][Discuss] Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000)…

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 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…

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…

poj 2391 Ombrophobic Bovines, 最大流, 拆点, 二分 dinic /* * Author: yew1eb * Created Time: 2014年10月31日 星期五 15时39分22秒 * File Name: poj2391.cpp */ #include <ctime> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring&g…

/** 题目:poj2391 Ombrophobic Bovines 链接:http://poj.org/problem?id=2391 题意:有n块区域,第i块区域有ai头奶牛,以及一个可以容纳bi头奶牛的棚子.n块区域由m条可以容纳无数奶牛经过的双向通道相连,给定奶牛通过通道的时间. 问所有奶牛回到棚子需要的最短时间. 思路:...我好菜哦.没想到,看了blog才知道怎么做. 先用floyd求得两块区域相通最短时间. 将点x拆分成x,x'.源点s连接x,容量为x的奶牛数.x'连接汇点t容量为…

[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…

bzoj1682[Usaco2005 Mar]Out of Hay 干草危机 题意: 给个图,每个节点都和1联通,奶牛要从1到每个节点(可以走回头路),希望经过的最长边最短. 题解: 求最小生成树即可. 代码: #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #define inc(i,j,k) for(int i=j;i<=k;i++) #defin…

bzoj1680[Usaco2005 Mar]Yogurt factory bzoj1740[Usaco2005 mar]Yogurt factory 奶酪工厂 题意: n个月,每月有一个酸奶需求量(吨)和酸奶成本(元每吨).酸奶可以保存,费用为S(元每月每吨),求最小总费用.n≤10000 题解: 第i月每吨酸奶的成本为Cj+s*(i-j),j∈[1,i],化简得Cj-s*j+s*i,因为s*i只和当前相关,所以维护一个最小的Cj-s*j即可.注意开long long. 代码: #includ…

二分答案,每一头牛向所有在规定时间内能走到的牛棚连inf的边,每一个源点向牛连牛数量的边,每一个牛棚向汇点连牛棚容量的边,能满流则意味着这个答案可行,否则不可行. 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 505 4 #define ll long long 5 #define inf 0x3f3f3f3f 6 struct ji{ 7 int nex,to,len; 8 }e[N*N],edge[N*N]; 9…

http://poj.org/problem?id=2391 floyd+网络流+二分 题意:有一个有向图,里面每个点有ai头牛,快下雨了牛要躲进雨棚里,每个点有bi个雨棚,每个雨棚只能躲1头牛.牛可以通过点之间的边移动来躲到其他的点的雨棚,路很宽一次可以走无数头牛,路的权是牛走过这条路要的时间.求最短需要多少时间所有牛都躲到雨棚下. 题解: 先floyd求出每个点到每个点的最短时间!然后二分答案,也就是二分牛躲完的时间,然后大于这个时间的路都封了,建个碉图,每个点分成两个点A和B,起点连接到所…

Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to cr…

[题意]有n块草地,一些奶牛在草地上吃草,草地间有m条路,一些草地上有避雨点,每个避雨点能容纳的奶牛是有限的,给出通过每条路的时间,问最少需要多少时间能让所有奶牛进入一个避雨点. 和POJ2112很类似,都是使最大的走过路径长度和最小的题,也是先Floyd再二分求最大流看是否满流--昨天刚做啊,第一下竟然没看出来,真是太年轻太Naive了!--=_= [思路]拆点建图,每个点的(in)和(out)间容量oo,源点向每个有牛的点(in)连边,容量为该点的牛数,每个有避雨点的点(out)向汇点连边,…

Description 牛们干草要用完了!贝茜打算去勘查灾情． 有N(2≤N≤2000)个农场,M(≤M≤10000)条双向道路连接着它们,长度不超过10^9．每一个农场均与农场1连通．贝茜要走遍每一个农场．她每走一单位长的路,就要消耗一单位的水．从一个农场走到另一个农场,她就要带上数量上等于路长的水．请帮她确定最小的水箱容量．也就是说,确定某一种方案,使走遍所有农场通过的最长道路的长度最小,必要时她可以走回头路． Input 第1行输入两个整数N和M;接下来M行,每行输入三个整数,表示一条道路…