命名冲突,导致编译失败.这大概就是之前看到的最好不要using namespace std:的原因…

题目如下: One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls…

这题写得比较痛苦.首先有点不在状态,其次题目比较难读懂. “Gang”成立的两个条件:①成员数大于两个  ②边权总和大于阈值K 首先,在录数据的时候通过map或者字符串哈希建立string到int的映射. 然后,这个题的数据结构其实是带权无向图.在录数据的时候就要处理好点权和边权. 最后,对所有顶点做一遍dfs,汇总边权,找出最大点权和最大点权所在的点,把数据录进ans里. 注意的点一个是环路边权的汇总.方法是在dfs中通过先加边权再判断vis是否遍历过.此时又要注意用不搜前驱点的dfs结果,及…

1034 Head of a Gang (30)(30 分) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time lengt…

1034. Head of a Gang (30) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of…

1034 Head of a Gang (30 分) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between Aand B, we say that A and B is related. The weight of a relation is defined to be the total time length of…

1034. Head of a Gang (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related.…

图的遍历的定义: 从图的某个顶点出发访问遍图中所有顶点,且每个顶点仅被访问一次.(连通图与非连通图) 深度优先遍历(DFS): 1.访问指定的起始顶点: 2.若当前访问的顶点的邻接顶点有未被访问的,则任选一个访问之:反之,退回到最近访问过的顶点:直到与起始顶点相通的全部顶点都访问完毕: 3.若此时图中尚有顶点未被访问,则再选其中一个顶点作为起始顶点并访问之,转 2: 反之,遍历结束. 连通图的深度优先遍历类似于树的先根遍历 如何判别V的邻接点是否被访问? 解决办法:为每个顶点设立一个“访问标志”…

图的存储结构 1)邻接矩阵 用两个数组来表示图,一个一维数组存储图中顶点信息,一个二维数组(邻接矩阵)存储图中边或弧的信息. 2)邻接表 3)十字链表 4)邻接多重表 5)边集数组 本文只用代码实现用邻接矩阵方式存储图.忘见谅. 图的遍历 1)深度优先遍历(Depth_First_Search,DFS) 从图中某个顶点 v 出发,访问此顶点,然后从 v 的未被访问的邻接点出发深度优先遍历图,直至图中所有和 v 有路径相通的顶点都被访问到.--------递归思想 2)广度优先遍历(Breadth…

一.图的遍历 #include<iostream> #include<queue> #include<vector> using namespace std; int n, m; //行数和列数 const int maxn = 100; char g[maxn][maxn]; //图 bool vis[maxn][maxn]; //访问标记数组,false表示点没有被访问过 int disx[4] = { 0,0,1,-1 }; //四个 int disy[4] =…