Tour Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3159    Accepted Submission(s): 1525 Problem Description In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000…

http://acm.hdu.edu.cn/showproblem.php?pid=3488 给一个无源汇的,带有边权的有向图 让你找出一个最小的哈密顿回路 可以用KM算法写,但是费用流也行 思路 1. 哈密顿回路对于每个点的流量有限制,因此$V$拆开为$V$和$V'$ 2. 我们建立附加源点$S$和附加汇点$T$哈密顿回路中的每个点有其唯一的后继和前驱,换句话说,对于任意一个点$V$,它满足$in(V)=out(V)$ 为了满足该条件,从源点向$V$ 连接容量为1,费用为0的边,从$V'$向汇…

题面 传送门 思路 仔细观察题目要求的东西,发现就是求一个最小路径覆盖,只不过可以跳跃(就是那个鬼畜的超级跳跃) 那么就直接上最小路径覆盖模版 对每个点,拆成两个点$X_i$和$Y_i$,建立超级源超级汇S,T 连边$\left(S,X_i\right)$,$\left(Y_i,T\right)$,流量1费用0 对于原图中的边$\left(i,j\right)$,连边$\left(X_i,Y_j\right)$,流量1费用为原本的时间 对于超级跳跃,连边$\left(S,Y_i\right)$,…

题目链接 BZOJ2597 题解 orz思维差 既然是一张竞赛图,我们选出任意三个点都可能成环 总方案数为 \[{n \choose 3}\] 如果三个点不成环,会发现它们的度数是确定的,入度分别为\(2,1,0\),出度为\(0,1,2\) 所以一个点的任意两个入度,都会对答案产生一个负的贡献 所以三元环数量为 \[{n \choose 3} - \sum\limits_{i = 1}^{n} {inde[i] \choose 2}\] 我们要最大化三元环数目,就要最小化\(\sum\limi…

Cyclic Tour Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 2399    Accepted Submission(s): 1231 Problem Description There are N cities in our country, and M one-way roads connecting them. Now L…

题意: 给你一个带权有向图,选择一些边组成许多没有公共边的环,使每个点都在k个环上,要求代价最小. SOL: 现在已经养成了这种习惯,偏题怪题都往网络流上想... 怎么做这题呢... 对我们看到每个点都在k个环上,而且没有公共边,那么很显然每个点的入度出度都为k.   然后我们拆点,建源汇ST,S与每个入点连边容量为k,出点与汇点相连容量为k,费用为0,如果城市i,j之间有边那么将i的入点和j的出点连一条费用为权,容量为1的边.然后跑一遍费用流.如果每条边都满流那么就有解. 好神奇...从环变成…

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…

http://poj.org/problem?id=2175 Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3256   Accepted: 855   Special Judge Description The City has a number of municipal buildings and a number of fallout shelters that were build…

题目链接 上下界费用流: /* 每个点i恰好(最少+最多)经过一次->拆点(最多)+限制流量下界(i,i',[1,1],0)(最少) 然后无源汇可行流 不需要源汇. 注: SS只会连i',求SS->TT的最大流 该走的i->i'是不会不走的 */ #include<queue> #include<cstdio> #include<cctype> #include<cstring> #include<algorithm> #def…

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 <…