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 made between the two persons. A "Gang" is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threshold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.
Input Specification:
Each input file contains one test case. For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threshold, respectively. Then N lines follow, each in the following format:
Name1 Name2 Time
where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.
Output Specification:
For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.
Sample Input 1:
8 59 AAA BBB 10 BBB AAA 20 AAA CCC 40 DDD EEE 5 EEE DDD 70 FFF GGG 30 GGG HHH 20 HHH FFF 10
Sample Output 1:
2 AAA 3 GGG 3
Sample Input 2:
8 70 AAA BBB 10 BBB AAA 20 AAA CCC 40 DDD EEE 5 EEE DDD 70 FFF GGG 30 GGG HHH 20 HHH FFF 10
Sample Output 2:
0
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<map> 5 #include<string> 6 using namespace std; 7 map<string , int> mp1; 8 map<int, string>mp2; 9 map<string, int> ans; 10 int G[2001][2001] = {0}, personNum = 0, time_[2001] = {0}; 11 int N, K, visit[2001] = {0}; 12 int str2num(string ss){ 13 if(mp1.count(ss) == 0){ 14 mp1[ss] = personNum; 15 mp2[personNum] = ss; 16 return personNum++; 17 }else return mp1[ss]; 18 } 19 void dfs(int v, int &sum, int &head, int &cnt){ 20 visit[v] = 1; 21 cnt++; 22 if(time_[head] < time_[v]) 23 head = v; 24 for(int i = 0; i < personNum; i++){ 25 if(G[v][i] != 0){ 26 sum += G[v][i]; 27 G[v][i] = G[i][v] = 0; 28 if(visit[i] == 0) 29 dfs(i, sum, head, cnt); 30 } 31 } 32 } 33 int main(){ 34 //std::ios::sync_with_stdio(false); 35 cin >> N >> K; 36 string str1, str2; 37 int we; 38 for(int i = 0; i < N; i++){ 39 cin >> str1 >> str2 >> we; 40 int e1 = str2num(str1); 41 int e2 = str2num(str2); 42 G[e1][e2] += we; 43 G[e2][e1] += we; 44 time_[e1] += we; 45 time_[e2] += we; 46 } 47 int gangs = 0; 48 for(int i = 0; i < personNum; i++){ 49 if(visit[i] == 0){ 50 int sum = 0, cnt = 0, head = i; 51 dfs(i, sum, head, cnt); 52 if(sum > K && cnt > 2){ 53 gangs++; 54 ans[mp2[head]] = cnt; 55 } 56 } 57 } 58 cout << gangs << endl; 59 map<string, int>::iterator it; 60 for(it = ans.begin(); it != ans.end(); it++){ 61 cout << it->first << " " << it->second << endl; 62 } 63 return 0; 64 }
总结：
1、题意：一个连通子图的节点数大于2，且它的边权的和大于K，就说明是gang，其中这个子图中与他人通话时长之和最多的节点被认为是头领。要求输出头领和该子图的节点个数。
2、首先是字符串作为节点的id需要被编号。起初看到只有3位想将其映射为26*26*26进制，但其实太大了而且有好多节点有可能是空的，这样从0遍历到26*26*26遍历的话时间上会超时。 所以只能使用map，映射从int到string，string到int。再使用一个计数器统一分配从0开始的节点编号。
3、由于深搜的原则是所有节点只遍历一次，使用了visit数组记录一个点是否被访问过。但本题实际上要求遍历连通图的所有边（获取整个gang的通话时长总和），如果仅仅按照点的方法遍历，很可能漏掉一些边。 所以
- 要在判断遍历A的所有联通的点是否被visit了之前就先访问一次他们之间的边。
- 其次为了防止回溯之后重复访问某条边，需要在累加完这条边之后立即置0。
第一种情况，深度搜索按照A->B->C的顺序访问节点，A时可以得到AB边，B时可以得到BC边，当访问C节点时，由于A被标记为visit，所以CA无法得到。这是回溯到A节点，A无法访问C（C被标记为visit），故AC边被漏掉了。
第二种情况，已经在判断visited之前加上了某条边，还会出重复访问某条边的情况。比如与之前类似，A->B->C的顺序访问，当访问到C时，由于这次是在判断visit之前先访问边，所以边CA被访问。结束后回溯到A，A再次测试C节点（不可访问，被标记visit），这是会再访问一次边AC，造成重复访问。
4、对于无向图，一定要对G[ i ][ j ]与G[ j ][ i ]同时操作。
5、本题判断谁是头目采用找点权最大的节点，这个可以在读入数据的时候就计算，不必等到遍历的时候。其次，由于是通话记录，所以可能会有多个A->B，B->A的记录，图上的边权是累加的而不是赋值（A->B需要给AB与BA边累加，B->A同样要给AB与BA累加）。
6、map<key, val>中，存储是按key递增的顺序记录的，遍历即可得到递增序列。
