01规划 设答案为 \(ans\). 二分答案,设当前二分值为 \(mid\). 设一个环 \(S\) 的边权为 \(t_1, t_2, t_3...\),点权为 \(f_1, f_2, f_3...\) 若 \(mid <= ans\),即存在一个环\(S\)使得 \(mid <= \frac{\sum f_i}{\sum t_i}\),变换一下:\(\sum(mid * t_i - f_i) <= 0\) 否则,则 \(mid > ans\) 每次 \(check\) 的时候,…

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…

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…

观光奶牛 农夫约翰已决定通过带他们参观大城市来奖励他们的辛苦工作!奶牛必须决定如何最好地度过他们的空闲时间. 幸运的是,他们有一个详细的城市地图,显示L(2≤L≤1000)主要地标(方便编号为1 .. L)和P(2≤P≤5000)单向奶牛路径加入它们.农夫约翰将把奶牛带到他们选择的起始地标,从那里他们将沿着牛路走到一系列其他地标,最后回到他们的起始地标,农民约翰将把他们捡起来带回农场.因为城市中的空间非常宝贵,所以奶牛路径非常狭窄,因此沿着每个奶牛路径行进仅允许在一个固定方向上行进. 虽然奶牛可…

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…

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…

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…

题意 题目链接 Sol 复习一下01分数规划 设\(a_i\)为点权,\(b_i\)为边权,我们要最大化\(\sum \frac{a_i}{b_i}\).可以二分一个答案\(k\),我们需要检查\(\sum \frac{a_i}{b_i} \geqslant k\)是否合法,移向之后变为\(\sum_{a_i} - k\sum_{b_i} \geqslant 0\).把\(k * b_i\)加在出发点的点权上检查一下有没有负环就行了 #include<bits/stdc++.h> #defin…

一道\(0/1\)分数规划+负环 POJ原题链接 洛谷原题链接 显然是\(0/1\)分数规划问题. 二分答案,设二分值为\(mid\). 然后对二分进行判断,我们建立新图,没有点权,设当前有向边为\(z=(x,y)\),\(time\)为原边权,\(fun\)为原点权,则将该边权换成\(mid\times time[z]+fun[x]\),然后在上面跑\(SPFA\). 如果有一个环使得\(\sum\{mid\times time[z]+fun[x]\}<0\),则说明\(mid\)小了,而式子…

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.…