Matrix Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2350 Accepted Submission(s): 1241 Problem Description Yifenfei very like play a number game in the n*n Matrix. A positive integer number…
最短路对应费用,路径数量对应流量.为限制点经过次数,拆点为边.跑一次流量为2的最小费用最大流. 最小费用最大流和最大流EK算法是十分相似的,只是把找增广路的部分换成了求费用的最短路. #include<bits/stdc++.h> using namespace std; typedef long long ll; +; +; struct Edge { int v,cap,cost,nxt; void IN(int V,int C,int c,int N) { v = V; cap = C;…
有一棵二叉树,如下图所示: 其中 # 表示空结点. 先序遍历:A B D E G C F 问题:怎么得到从根结点到任意结点的路径呢? 示例:输入 G,怎么得到从结点 A 到结点 G 的路径呢? 很明显,我们一眼就能看出来路径是 A B E G.如何通过程序得到这条路径就是我们接下来需要做的. 定义二叉树的 链式存储结构 如下: typedef struct BiTNode { char data; struct BiTNode* lchild, * rchild; }BiTNode, * BiT…
最小割最大流定理:(参考刘汝佳p369)增广路算法结束时,令已标号结点(a[u]>0的结点)集合为S,其他结点集合为T=V-S,则(S,T)是图的s-t最小割. Problem Description You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD 1 fr…
ACM Computer Factory Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10940 Accepted: 4098 Special Judge Description As you know, all the computers used for ACM contests must be identical, so the participants compete on equal terms. T…
