题目链接:http://poj.org/problem?id=3621 Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10552   Accepted: 3613 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big ci…

题目链接:id=3621">http://poj.org/problem?id=3621 在一个有向图中选一个环,使得环上的点权和除以边权和最大.求这个比值. 经典的分数规划问题,我认为这两篇题解写得非常清楚,能够參考一下http://blog.csdn.net/gengmingrui/article/details/47443705,http://blog.csdn.net/wall_f/article/details/8221807 个人认为01分数规划差点儿都是二分或者迭代比值,从而…

感觉去年9月的自己好$naive$ http://www.cnblogs.com/candy99/p/5868948.html 现在不也是嘛 裸题,具体看学习笔记 二分答案之后判负环就行了 $dfs$版超快 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; typed…

Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8915   Accepted: 3000 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to…

题意: 给定n个点,每个点有一个开心度F[i],每个点有m条单向边,每条边有一个长度d,要求一个环,使得它的 开心度的和/长度和 这个比值最大.n<=1000,m<=5000 题解: 最优比率环,很像以前做过的一题最优比率生成树.首先二分一个答案r=sigma(xi*fi)/sigma(xi*di),设z=r*sigma(xi*di)-sigma(xi*fi),若能找到一个环满足z<=0,则代表sigma(xi*fi)>=r*sigma(xi*di),则r可以比当前的r要大,则l=…

01分数规划 前置技能 二分思想最短路算法一些数学脑细胞? 问题模型1 基本01分数规划问题 给定nn个二元组(valuei,costi)(valuei,costi),valueivaluei是选择此二元组获得的价值(非负),costicosti是选择此二元组付出的代价(非负),设xi(xi∈{0,1})xi(xi∈{0,1})代表第ii个二元组的选与不选,最大(小)化下式 maximize(or minimize)   r=∑valuei⋅xi∑costi⋅ximaximize(or mini…

题目链接:http://poj.org/problem?id=3621 思路:之前做过最小比率生成树,也是属于0/1整数划分问题,这次碰到这道最优比率环,很是熟悉,可惜精度没控制好,要不就是wa,要不就是tle,郁闷啊!实在是懒得码字,直接copy吧: 题目的意思是:求一个环的{点权和}除以{边权和},使得那个环在所有环中{点权和}除以{边权和}最大.令在一个环里,点权为v[i],对应的边权为e[i], 即要求:∑(i=1,n)v[i]/∑(i=1,n)e[i]最大的环(n为环的点数), 设题目…

最优比率环问题.二分答案,对于每一个mid,把节点的happy值归类到边上. 对于每条边,用mid×weight减去happy值,如果不存在负环,说明还可以更大. /*--------------------------------------------------------------------------------------*/ // Helica's header // Second Edition // 2015.11.7 // #include <algorithm> #i…

题意: 给定L个点, P条边的有向图, 每个点有一个价值, 但只在第一经过获得, 每条边有一个花费, 每次经过都要付出这个花费, 在图中找出一个环, 使得价值之和/花费之和 最大 分析: 这道题其实并不是很好想, 因为价值和花费不是在同一样东西, 价值是点, 花费是边. 但回到我们要求的问题上, 我们要找出一个最优比率的环, 那么其实每个点只会经过一次, 是一个单独的环, 所以我们可以把价值也视为边的一部分. 参考这篇博客http://blog.csdn.net/gengmingrui/arti…

http://poj.org/problem?id=3621 题意:有n个点m条有向边,每个点有一个点权val[i],边有边权w(i, j).找一个环使得Σ(val) / Σ(w)最大,并输出. 思路:和之前的最优比率生成树类似,还是构造成这样的式子:F(L) = Σ(val[i] * x[i]) - Σ(w[i] * x[i] * L) = Σ(d[i]) * x[i] (d[i] = val[i] - w[i] * L),要使得L越大越好. 那么当L越大的时候,F(L)就越小,如果F(L)大…

Sightseeing Cows Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time. Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000)…

典型的求最优比例环问题 參考资料: http://blog.csdn.net/hhaile/article/details/8883652 此题中,给出每一个点和每条边的权值,求一个环使 ans=∑点权/∑边权 最大. 由于题目要求一个环,并且必定是首尾相接的一个我们理解的纯粹的环,不可能是其它样子的环, 所以我们能够把一条边和指向的点看做总体处理. 上面方程能够化为:ans×e[i]-p[i]=0 以它为边权二分答案,spfa求负环,有负环则该ans可行,增大下界. 若一直不可行,则无解. #…

题目大意:在一个无向图里找一个环,是的点权和除以边权和最大 思路:UVA11090姊妹题 事实上当这题点权和都为1时就是上一题TUT #include <stdio.h> #include <iostream> #include<queue> #include <string.h> #include <algorithm> #define maxn 10009 #define maxm 10001 #define esp 0.001 using…

01分数规划 二分+spfa负环(SLF优化) #include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> using namespace std; int n,m; ]; ]; struct node { int x,y,next;double d; }a[];]; void i…

Sightseeing Cows 给出一张图,点数为L,边数P,并给出边的边权\(\{b_i\}\),再给处每个点的点权,求一条起点和终点相同的路径,并使其点权之和除以边权之和最大,注意,路径中点权只能被计算一次,而边权可以重复计算, (2 ≤ L ≤ 1000), (2 ≤ P ≤ 5000). 解 显然为分数规划问题,关键在点权与边权不对应上,于是自然的想法是点权移边权,而一条起点与终点相同的路径即一个联通分量,所以问题现在在于点权移边权后只对环成立,而不对联通分量成立,于是考虑证明联通分量…

题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time. Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landma…

01分数规划复习. 这东西有一个名字叫做最优比率环. 首先这个答案具有单调性,我们考虑如何检验. 设$\frac{\sum_{i = 1}^{n}F_i}{\sum_{i = 1}^{n}T_i} = e$,我们需要检验的就是$\sum_{i = 1}^{n}(F_i - mid * T_i) \geq 0$是否存在. 感觉这玩意不好算,再变形一下:$\sum_{i = 1}^{n}(e * T_i - F_i) < 0$,就变成一个负环的检验了. $F_i$应当可以任取一条有向边的入点和出点.…

[题目链接] http://poj.org/problem?id=3621 [算法] 01分数规划(最优比率环) [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio&g…

题目描述: David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be water…

Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10306   Accepted: 3519 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to…

[POJ3621]Sightseeing Cows 题意:在给定的一个图上寻找一个环路,使得总欢乐值(经过的点权值之和)/ 总时间(经过的边权值之和)最大. 题解:显然是分数规划,二分答案ans,将每条边的权值变成(ans*边权-2*起始点点权),然后我们希望找出一个环,使得环上的总边权<0 (一开始我把题意理解错了,题中给的是单向边,我把它当成是双向边+每条边只能走一次了~,想出一堆做法都接连pass掉) 然后就直接用SPFA判负环就好了嘛!由于原图不一定联通,所以一开始就把所有点都入队就完事…

[POJ3621][洛谷2868]Sightseeing Cows(分数规划) 题面 Vjudge 洛谷 大意: 在有向图图中选出一个环,使得这个环的点权\(/\)边权最大 题解 分数规划 二分答案之后把每条边的边权换为\(mid·\)边权-出点的点权 然后检查有没有负环就行啦 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath&g…

Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8331   Accepted: 2791 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to…

题目链接:http://poj.org/problem?id=3621 Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11526   Accepted: 3930 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big ci…

[USACO07DEC]Sightseeing Cows Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time. Fortunately, they have a detailed city map showi…

P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time. Fortunately, they have a detailed city map sh…

题外话:最近差不多要退役,复赛打完就退役回去认真读文化课. 题面:P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题解:最优比例环 题目实际是要求一个ans,使得对于图中任意一个环满足 sig(i=1,n)v[i]/sig(i=1,n)e[i]<=ans 所以将公式变换为:sig(i=1,n)v[i]-[(sig(i=1,n)v[i])*ans]<=0 sig(i=1,n)(v[i]-ans*e[i])<=0 最终化为:sig(i=1,n)(ans*e[i]…

题意描述 Sightseeing Cows G 给定一张有向图,图中每个点都有点权 \(a_i\),每条边都有边权 \(e_i\). 求图中一个环,使 "环上个点权之和" 除以 "环上各边权之和" 最大.输出最大值. 解释一下,原题目中并没有点明这是一个环,但是从: 奶牛们不会愿意把同一个建筑物参观两遍. 可以看出,不管怎么走,走环一定是最优的,因为重复走相当于无故增加分母. 算法分析 据说这是一道 0/1 分数规划的题目,但是其实可以用更加通俗易懂的方法来解释.…

P2868 [USACO07DEC]观光奶牛Sightseeing Cows [](https://www.cnblogs.com/images/cnblogs_com/Tony-Double-Sky/1270353/o_YH[_INPMKE_4RY]3DF(33@G.png) 错误日志: dfs 判负环没有把初值赋为 \(0\) 而是 \(INF\), 速度变慢 Solution 设现在走到了一个环, 环内有 \(n\) 个点, \(n\) 条边, 点权为 \(f_{i}\), 边权为 \(e…

嘟嘟嘟 这题好像属于01分数规划问题,叫什么最优比率生成环. 题目概括一下,就是求一个环,满足∑v[i] / ∑c[i]最大. 我们可以堆上面的式子变个型:令 x = ∑v[i] / ∑c[i],则x * ∑c[i] = v[i] => ∑x * c[i] - v[i] = 0.于是对于任何能取到的x',满足∑x * c[i] - v[i] <= 0:对于不能取到的x', ∑x * c[i] - v[i] > 0. 于可以实数二分答案,用spfa判断负环. 然后实数二分又RE了……调了好…