01的世界

题目：hdu 5438PondsTime Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 76    Accepted Submission(s): 28Problem DescriptionBetty owns

Play on WordsSome of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the pu

图论中的优先级搜索——DFS,BFS,Prim,Dijkstra  在图算法中经常要执行遍历每个顶点和每条边的操作，即图搜索。许多图算法都以图搜索为基础，如2-着色问题、连通性计算基于深度优先搜寻（depth-first search, DFS），而无权最短路径则基于广度优先搜索（breadth-first search， BFS）。基于搜索的算法还包括计算最小生成树的Prim算法

Network of SchoolsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13328 Accepted: 5330DescriptionA number of schools are connected to a computer network.

POJ 2186 Popular Cows（强连通分量）Popular CowsTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13835 Accepted: 5484DescriptionEvery cow's dream is to become

The Bottom of a GraphTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 7098 Accepted: 2905DescriptionWe will use the following (standard) definitions from

题意：有n个人m通电话，如果有两个人相互打电话(直接或间接)则在同一个电话圈里。输出所有电话圈的人的名单。分析：用floyd求出两点之间是否有边，然后如果g[i][j]==g[j][i]==1，那么就放入一个连通分量，最后依次输出每个连通分量的所有边，注意输入输出格式，这地方坑了几次。放入一个连通分量用并查集做。每条边对应一个ID号用map去重。找连通分量lrj是用的dfs