题意:2个人从1走到n,假设一条路第一次走则是价值di,假设第二次还走这条路则须要价值di+ai,要你输出2个人到达终点的最小价值! 太水了!一条边建2次就OK了.第一次价值为di,第二次为ai+di,加入源点汇点跑最小费用最大流就OK了! AC代码: #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #includ…
描述 Alice and Bob are walking in an ancient maze with a lot of caves and one-way passages connecting them. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are too small for two people to walk through simultaneou…
时间限制:0.25s 空间限制:4M 题意: 在n(n<=400)个点的图中,找到并输出两条不想交的最短路.不存在输出“No sulotion”: Solution: 最小费用最大流 建图与poj 2135 一样,添加S到1的流量为2权为0,n到T的流量为2权为0的边,其它边的流量为1,权为路径长度. 但是这道题麻烦不在要输出最短路,而在仅仅4M的内存上. 由于只有4M,我们最多存上400*400条边.但是图却是一个无向图,朴素的想法是存上400*400*2条边,但是这里内存不够. 所以我们首先…
[原题](http://poj.org/problem?id=3068) 给一个有向带权图,求两条从0-N-1的路径,使它们没有公共点且边权和最小 . //是不是像传纸条啊- 是否可行只要判断最后最大流是不是2就可以了 #include<cstdio> #include<queue> #include<cstring> #define N 1010*1010 #define inf 0x3f3f3f3f using namespace std; int n,m,head…
题目求一张图两条边不重复的最短路. 一开始我用费用流做. 源点到1连容量2费用0的边:所有边,连u到v和v到u容量1费用cost的边. 总共最多会增广两次,比较两次求得的费用,然后输出路径. 然而死MLE不过.. 看了题解,是用最大流的做的. 源点到1连容量为2的边:然后把属于最短路的边都加进去,容量为1. 跑一遍最大流,如果流量为2,那就有解,最后再从1到n沿着满流的边输出两条路径. 学到了怎么求出所有属于最短路的边... #include<cstdio> #include<cstri…
Libre 6013 「网络流 24 题」负载平衡 (网络流,最小费用最大流) Description G 公司有n 个沿铁路运输线环形排列的仓库,每个仓库存储的货物数量不等.如何用最少搬运量可以使n 个仓库的库存数量相同.搬运货物时,只能在相邻的仓库之间搬运. «编程任务: 对于给定的n 个环形排列的仓库的库存量,编程计算使n 个仓库的库存数量相同的最少搬运量. Input 第1 行中有1 个正整数n(n<=100),表示有n个仓库. 第2 行中有n个正整数,表示n个仓库的库存量. Outpu…
Libre 6011 「网络流 24 题」运输问题 (网络流,最小费用最大流) Description W 公司有m个仓库和n个零售商店.第i个仓库有\(a_i\)个单位的货物:第j个零售商店需要\(b_j\)个单位的货物.货物供需平衡.从第i个仓库运送每单位货物到第j个零售商店的费用为\(c_{ij}\).试设计一个将仓库中所有货物运送到零售商店的运输方案,使总运输费用最少. Input 第1行有2个正整数m和n,分别表示仓库数和零售商店数.接下来的一行中有m个正整数\(a_i\),表示第i个…
Libre 6008 「网络流 24 题」餐巾计划 (网络流,最小费用最大流) Description 一个餐厅在相继的N天里,第i天需要Ri块餐巾(i=l,2,-,N).餐厅可以从三种途径获得餐巾. (1)购买新的餐巾,每块需p分: (2)把用过的餐巾送到快洗部,洗一块需m天,费用需f分(f<p).如m=l时,第一天送到快洗部的餐巾第二天就可以使用了,送慢洗的情况也如此. (3)把餐巾送到慢洗部,洗一块需n天(n>m),费用需s分(s<f). 在每天结束时,餐厅必须决定多少块用过的餐巾…
1506: Double Shortest Paths Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 49  Solved: 5 Description Input There will be at most 200 test cases. Each case begins with two integers n, m (1<=n<=500, 1<=m<=2000), the number of caves and passages.…
传送门:Double Shortest Paths 题意:有两个人:给出路径之间第一个人走所需要的费用和第二个人走所需要的费用(在第一个人所需的 费用上再加上第二次的费用):求两个人一共所需要的最小费用. 分析:建立超源和超汇,流量分别为2,从源点到汇点的最大流2时最小费用为答案. #include <cstdio> #include <cstring> #include <string> #include <queue> #include <cmat…
Double Shortest PathsAlice and Bob are walking in an ancient maze with a lot of caves and one-way passages connectingthem. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are toosmall for two people to walk thr…
题目链接:http://acm.hit.edu.cn/hoj/problem/view?id=2739 Time limit : 1 sec Memory limit : 64 M A Chinese postman is assigned to a small town in China to deliver letters. In this town, each street is oriented and connects exactly two junctions. The postma…
题目大概说给一张有向图,要从0点出发返回0点且每条边至少都要走过一次,求走的最短路程. 经典的CPP问题,解法就是加边构造出欧拉回路,一个有向图存在欧拉回路的充分必要条件是基图连通且所有点入度等于出度. 而这题,果断联想到混合图欧拉回路的做法,用最小费用最大流解决: 先只考虑所有边都只走一次,计算出各个点的出度和入度,出度不等于入度的点就需要选择几条边去改变调整它们 对于出度多的就和容量网络的汇点连容量出度-入度费用0的边,入度多的源点就向其同样地连边 对于原图中的所有边<u,v>由u向v连容…
题意:M个影片,其属性有开始时间S,结束时间T,类型op和权值val.有K个人,每个人可以看若干个时间不相交的影片,其获得的收益是这个影片的权值val,但如果观看的影片相邻为相同的属性,那么收益要减少W.每个影片只能被一个人看.求所有人能获得的收益值之和的最大值. 分析:因为人数不定,所以贪心和dp的思路被否定了.1对多的带权匹配,求最大权,这种问题显然KM是解决不了的,那么只能是最小费用最大流了.而这题要求的是最大收益,那么建负权边即可. 为了保证每个影片只被一个人观看,将其拆为入点和出点,入…
Farm Tour Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: 2135 64-bit integer IO format: %lld      Java class name: Main   When FJ's friends visit him on the farm, he likes to show them around. His farm compr…
Coding Contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1751    Accepted Submission(s): 374 Problem Description A coding contest will be held in this university, in a huge playground. The…
POJ 2135 Farm Tour (网络流,最小费用最大流) Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the b…
题目链接:http://poj.org/problem?id=2135 Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17672   Accepted: 6851 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17230   Accepted: 6647 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
Coding Contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2653    Accepted Submission(s): 579 Problem Description A coding contest will be held in this university, in a huge playground. The…
裸的费用流.往返就相当于从起点走两条路到终点. 按题意建图,将距离设为费用,流量设为1.然后增加2个点,一个连向节点1,流量=2,费用=0;结点n连一条同样的弧,然后求解最小费用最大流.当且仅当最大流=2时,有solution,此时费用即answer. -------------------------------------------------------------------------------- #include<cstdio> #include<cstring>…
题目链接:https://www.nowcoder.com/acm/contest/207/G 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5988 时间限制:C/C++ 2秒,其他语言4秒空间限制:C/C++ 262144K,其他语言524288K64bit IO Format: %lld题目描述 A coding contest will be held in this university, in a huge playground. Th…
描述 When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000)…
Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18150   Accepted: 7023 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13509   Accepted: 5125 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
度度熊的交易计划 Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1111    Accepted Submission(s): 403 Problem Description 度度熊参与了喵哈哈村的商业大会,但是这次商业大会遇到了一个难题: 喵哈哈村以及周围的村庄可以看做是一共由n个片区,m条公路组成的地区. 由于生产能力的区别,第i…
Farm Tour Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19207   Accepted: 7441 Description When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of…
http://poj.org/problem?id=3686 题意:给出n个玩具和m个工厂,每个工厂加工每个玩具有一个时间,问要加工完这n个玩具最少需要等待的平均时间.例如加工1号玩具时间为t1,加工2号玩具时间为t2.那么先加工玩具1再加工玩具2花费的时间是t1+(t1+t2),先加工玩具2在加工玩具1花费的时间是t2+(t1+t2). 思路:假设所有玩具在一个工厂加工,那么等待的时间是 t1 + (t1 + t2) + (t1 + t2 + t3) + …… = t1 * n + t2 *…
题意: 有 n+1 个城市编号 0..n,有 m 条无向边,在 0 城市有个警察总部,最多可以派出 k 个逮捕队伍,在1..n 每个城市有一个犯罪团伙,          每个逮捕队伍在每个城市可以选择抓或不抓,如果抓了 第 i  个城市的犯罪团伙,第 i-1 个城市的犯罪团伙就知道了消息  ,如果第 i-1 的犯罪 团伙之前没有被抓,任务就失败,问要抓到所有的犯罪团伙,派出的队伍需要走的最短路是多少. 分析: 最小费用最大流,需要注意的地方在于怎么去保证每个每个城市的团伙仅仅被抓一次,且在抓他…