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…

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…

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 **************************************************************…

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…

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…

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…

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…

Description A crime has been comitted: a load of grain has been taken from the barn by one of FJ's cows. FJ is trying to determine which of his C (1 <= C <= 100) cows is the culprit. Fortunately, a passing satellite took an image of his farm M (1 &l…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1680 题意: 在接下来的n周内,第i周生产一吨酸奶的成本为c[i],订单为y[i]吨酸奶. 酸奶可以提前生产,可以存放无限长的时间,存放一周的花费为s. 问你在完成所有订单的前提下,最小的花费为多少. 题解: 贪心. p[i]代表第i周的最小成本. 对于p[i],只用考虑p[i-1],因为已经保证了p[i-1]是i-1之前所有周的最优答案. 所以转移为:p[i] = min(c[i],…

贪心,一边读入一边更新mn,用mn更新答案,mn每次加s #include<iostream> #include<cstdio> using namespace std; int n,s,mn=1e9; long long ans; int main() { scanf("%d%d",&n,&s); for(int i=1,w,c;i<=n;i++) { scanf("%d%d",&w,&c); mn+=…

二分答案,把边权小于mid的边的两端点都并起来,看最后是否只剩一个联通块 #include<iostream> #include<cstdio> using namespace std; const int N=2005; int n,m,f[N]; struct qwe { int u,v,w; }a[N*5]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f…

Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) c…

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>=…

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…

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)…

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…

BZOJ 1584: [Usaco2009 Mar]Cleaning Up 打扫卫生 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 419  Solved: 278 Description 有N头奶牛,每头那牛都有一个标号Pi,1 <= Pi <= M <= N <= 40000.现在Farmer John要把这些奶牛分成若干段,定义每段的不河蟹度为:若这段里有k个不同的数,那不河蟹度为k*k.那总的不河蟹度就是所有段的不河蟹度的总和…

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…

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…

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…