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…

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…

http://www.lydsy.com/JudgeOnline/problem.php?id=1682 最小生成树裸题.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std;…

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…

题目描述 Bessie 计划调查N (2 <= N <= 2,000)个农场的干草情况,它从1号农场出发.农场之间总共有M (1 <= M <= 10,000)条双向道路,所有道路的总长度不超过1,000,000,000.有些农场之间存在着多条道路,所有的农场之间都是连通的. Bessie希望计算出该图中最小生成树中的最长边的长度. 输入格式 两个整数N和M. 接下来M行,每行三个用空格隔开的整数A_i, B_i和L_i,表示A_i和 B_i之间有一条道路长度为L_i. 输出格式…

...最小生成树裸题,9月最后一天刷水刷水. #include<iostream> #include<cstdio> #include<algorithm> using namespace std; +; +; struct Rec { int ori,des,len; bool operator < (const Rec &x) const { return len<x.len; } }edge[MAXM]; int par[MAXN],heigh…

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…

题目 1615: [Usaco2008 Mar]The Loathesome Hay Baler麻烦的干草打包机 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John新买的干草打包机的内部结构大概算世界上最混乱的了,它不象普通的机器一样有明确的内部传动装置,而是,N (2 <= N <= 1050)个齿轮互相作用,每个齿轮都可能驱动着多个齿轮. FJ记录了对于每个齿轮i,记录了它的3个参数:X_i,Y_i表示齿轮中心的位置…

1615: [Usaco2008 Mar]The Loathesome Hay Baler麻烦的干草打包机 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 574  Solved: 226[Submit][Status] Description Farmer John新买的干草打包机的内部结构大概算世界上最混乱的了,它不象普通的机器一样有明确的内部传动装置,而是,N (2 <= N <= 1050)个齿轮互相作用,每个齿轮都可能驱动着多个齿轮. FJ…

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)…

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…

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…

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…

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…

Description Farmer John新买的干草打包机的内部结构大概算世界上最混乱的了,它不象普通的机器一样有明确的内部传动装置,而是,N (2 <= N <= 1050)个齿轮互相作用,每个齿轮都可能驱动着多个齿轮. FJ记录了对于每个齿轮i,记录了它的3个参数:X_i,Y_i表示齿轮中心的位置坐标(-5000 <= X_i <= 5000; -5000 <= Y_i <= 5000):R_i表示该齿轮的半径(3 <= R_i <= 800).驱动…

http://www.lydsy.com/JudgeOnline/problem.php?id=1615 这种题..... #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std;…

直接bfs即可,注意开double,还有驱动和终点的齿轮都在序列里,要把它们找出来= = #include<iostream> #include<cstdio> #include<algorithm> #include<queue> using namespace std; const int N=1505; int n,sx,sy,s,t,q[N],fr[N]; bool v[N]; double ans,dis[N]; struct qwe { int…

为了调整电灯亮度,贝西要用干草包堆出一座塔,然后爬到牛棚顶去把灯泡换掉.干草包会从传送带上运来,共会出现N包干草,第i包干草的宽度是W i ,高度和长度统一为1.干草塔要从底层开始铺建.贝西会选择最先送来的若干包干草,堆在地上作为第一层,然后再把紧接着送来的几包干草包放在第二层, 再铺建第三层--重复这个过程, 一直到所有的干 草全部用完.每层的干草包必须紧靠在一起,不出现缝隙,而且为了建筑稳定,上层干草的宽度不能超过下层的宽度. 按顺序运来的干草包一定要都用上, 不能将其中几个干草包弃置不用.…

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

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 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…

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…

http://www.lydsy.com/JudgeOnline/problem.php?id=1681 太裸了.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std; #de…

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…

http://www.lydsy.com/JudgeOnline/problem.php?id=1680 看不懂英文.. 题意是有n天,第i天生产的费用是c[i],要生产y[i]个产品,可以用当天的也可以用以前的(多生产的).每单位产品保存一天的费用是s.求最小费用 显然贪心,每次查找之前有没有哪一天保存到现在的价值最小,然后比较更新.. #include <cstdio> #include <cstring> #include <cmath> #include <…

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],…

n<=400个东西,每个东西有高度<=100,这种东西在堆放过程中不得超过的最大高度<=40000,以及每个东西的个数<=10,求最高能堆多高. 算了下背包复杂度不太对然后开了bitset.. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> #include<bitset> //#include<iostream&…

n<=10000天每天Ci块生产一东西,S块保存一天,每天要交Yi件东西,求最少花多少钱. 这个我都不知道归哪类了.. #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> //#include<iostream> using namespace std; int n,m; #define maxn 10011 int a,b; #define…

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+=…

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…