Aizu 2304 Reverse Roads 费用流】的更多相关文章

Reverse Roads Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93265#problem/E Description ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the c…
把有向图修改成无向图,并保证每条边的流量守恒并满足有向容量(即abs(flow(u,v) - flow(v,u)) <= 1)满足限制. 得到最大流,根据残流输出答案. 因为最后少了'\n'而WA... #include<bits/stdc++.h> using namespace std; ,M = N*(N-); int n,m; int hd[N],nx[M],to[M],cap[M],ect; inline ) { nx[ect] = hd[u]; to[ect] = v; ca…
原题链接:http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2304 题意: 给你一个网络,其中每条边的容量是1,你可以通过调整边的方向来获得更大的流量,现在问你能获得的最大流量是多少.并且输出更改方向的边的编号. 题解: 就每条边弄成无向的,并且标记一下是否是原始边,然后跑一发Dinic即可.然后在残余网络上寻找解即可. 代码: #include<iostream> #include<stack> #include&…
Description In a country there are n cities connected by m one way roads. You can paint any of these roads. To paint a road it costs d unit of money where d is the length of that road. Your task is to paint some of the roads so that the painted roads…
3638: Cf172 k-Maximum Subsequence Sum Time Limit: 50 Sec  Memory Limit: 256 MBSubmit: 174  Solved: 92[Submit][Status][Discuss] Description 给一列数,要求支持操作: 1.修改某个数的值 2.读入l,r,k,询问在[l,r]内选不相交的不超过k个子段,最大的和是多少. Input The first line contains integer n (1 ≤ n …
Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1399    Accepted Submission(s): 712 Problem Description There are N cities in our country, and M one-way roads connecting them. Now Li…
Description There are N cities, and M directed roads connecting them. Now you want to transport K units ofgoods from city 1 to city N. There are many robbers on the road, so you must be very careful. Themore goods you carry, the more dangerous it is.…
题目描述 给一列数,要求支持操作: 1.修改某个数的值 2.读入l,r,k,询问在[l,r]内选不相交的不超过k个子段,最大的和是多少. 输入 The first line contains integer n (1 ≤ n ≤ 105), showing how many numbers the sequence has. The next line contains n integers a1, a2, ..., an (|ai| ≤ 500). The third line contain…
昨天考试被教育了一波.为了学习一下\(T3\)的科技,我就找到了这个远古时期的\(cf\)题(虽然最后\(T3\)还是不会写吧\(QAQ\)) 顾名思义,这个题目其实可以建成一个费用流的模型.我们用流量来限制区间个数,用费用强迫它每次每次选择最大的区间就可以啦.但是因为询问很多,复杂度似乎不行,于是就有了这种神奇的科技--线段树模拟费用流. 在原先的费用流模型里,我们有正反两种边,而反向边的意义就在于,在每一次增广的时候可以反悔以前的操作,把局部最优向更大范围的局部更优优化. 参考反向边的原理,…
洛谷 Codeforces bzoj1,bzoj2 这可真是一道n倍经验题呢-- 思路 我首先想到了DP,然后矩阵,然后线段树,然后T飞-- 搜了题解之后发现是模拟费用流. 直接维护选k个子段时的最优解似乎也可以做,然而复杂度是O(nk2logn),显然跑不过. 考虑一种费用流做法.序列里每个点拆成入点和出点,源连入汇连出,入点和出点间连流量1费用ai的边,相邻点出点向入点连流量1费用0的边,整体限流k. 直接跑当然还不如暴力.观察一下这个做法是在干啥:每次选择费用最大的一段,然后利用反向边将这…