题目链接: http://codeforces.com/problemset/problem/546/E E. Soldier and Traveling time limit per test1 secondmemory limit per test256 megabytes 问题描述 In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of…

CF546E Soldier and Traveling 题目描述 In the country there are \(n\) cities and \(m\) bidirectional roads between them. Each city has an army. Army of the i-th city consists of \(a_{i}\) soldiers. Now soldiers roam. After roaming each soldier has to eith…

B. Soldier and Traveling Time Limit: 1000ms Memory Limit: 262144KB 64-bit integer IO format: %I64d      Java class name: (Any) Submit Status In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the …

In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the i-th city consists of ai soldiers. Now soldiers roam. After roaming each soldier has to either stay in his city or to go to the one of neighb…

题目描述 In the country there are n n n cities and m m m bidirectional roads between them. Each city has an army. Army of the i i i -th city consists of ai a_{i} ai​ soldiers. Now soldiers roam. After roaming each soldier has to either stay in his city o…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the i-th city consists of ai soldiers. N…

题目大概说一张无向图,各个结点初始有ai人,现在每个人可以选择停留在原地或者移动到相邻的结点,问能否使各个结点的人数变为bi人. 如此建容量网络: 图上各个结点拆成两点i.i' 源点向i点连容量ai的边 i'向汇点连容量bi的边 i向i'连容量INF的边 对于相邻的两点(u,v),u向v'连容量INF的边,v'向u连容量INF的边 跑最大流看看最大流是否等于∑bi.另外,注意∑ai不等于∑bi的情况. #include<cstdio> #include<cstring> #incl…

基础网络流,增加s和t,同时对于每个结点分裂为流入结点和流出结点.EK求最大流,判断最大流是否等于当前总人数. /* 304E */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #in…

题目链接 给出n个城市, 以及初始时每个城市的人数以及目标人数.初始时有些城市是相连的. 每个城市的人只可以待在自己的城市或走到与他相邻的城市, 相邻, 相当于只能走一条路. 如果目标状态不可达, 输出no, 否则输出每个城市的人都是怎么走的, 比如第一个城市有2个人走到了第二个城市, 1个人留在了第一个城市, 那么输出的第一行前两个数就是1, 2. 很明显的网络流, 输出那里写了好久... 首先判断能否可达, 如果初始状态的人数与目标状态不一样, 一定不可达, 其次, 如果建完图跑网络流的结果…

题意 (CodeForces 546E) 对一个无向图,给出图的情况与各个节点的人数/目标人数.每个节点的人只可以待在自己的城市或走到与他相邻的节点. 问最后是否有解,输出一可行解(我以为是必须和答案一样,然后本机调了半天死活不一样,交上去结果A了- -). 分析 典型的网络流.问题在于建模.如何解决两个节点的人数->目标人数? 考虑到我们始终要和这两个状态打交道,不妨将每个结点拆成两个(转移前&转移后),这两个点间的流量是INF.如果两点相连(不妨设为u,u',v,v'),那么uu与v′v…