1034. Head of a Gang (30)-PAT甲级真题

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 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 threthold, 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

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　　分析：题目大意是这样的，一帮人打电话，给你代号和代号之间的通话时间，谁自己的电话时间长就是那个组织的boss，然后如何判定一个组织，就是两个人以上且通话时间大于k这个值就算一个组织。然后输出boss，和组织人数。

　　实际上每一个连通分量只要边权之和大于k就是一个组织，如何求连通分量就是dfs，一切解决。

#include <iostream>#include <map>#include <string>using namespace std;map<string , int> nameToId;//代号和序号对应的图map<int , string> idToName;//序号和代号对应的图map<string , int> result;//最后输出的结果图int number = 1 , k;//number记录总人数，以及用于定义任务序号int transform (string s){    //如果这个人的代号不存在，那么加入这个人进入两个对应的图，设置上序号    if (nameToId[s] == 0)    {        nameToId[s] = number;        idToName[number] = s;        return number++;    }    //如果这个人的代号比如BBB已经存在,那么就返回他的序号    else    {        return nameToId[s];    }}int G[2017][2017] , weight[2017];//G为边权，weight为点权bool vis[2017];//深度优先遍历void dfs (int u , int &head , int &numMember , int &totalweight){    vis[u] = true;    numMember++;    //如果当前的人比boss的点权大那就重新定义谁是boss    if (weight[u] > weight[head])        head = u;    //当前遍历的人为u，查找他与其他的联系    for (int v = 1; v < number; v++)    {        //如果有联系，也就是是一个新的连通分量就记录        if (G[u][v] > 0)        {            totalweight += G[u][v];            G[u][v] = G[v][u] = 0;            if (vis[v] == false)                //继续遍历v                dfs (v , head , numMember , totalweight);        }    }}//主要的函数呵呵void dfsTrave (){    for (int i = 1; i < number; i++)    {        //一个个找连通分量去        if (vis[i] == false)        {            //假定当前人是boss，然后dfs            int head = i , numMember = 0 , totalweight = 0;            dfs (i , head , numMember , totalweight);            if (numMember > 2 && totalweight > k)                result[idToName[head]] = numMember;        }    }}int main (){    int n , w;    cin >> n >> k;    string s1 , s2;    for (int i = 0; i < n; i++)    {        cin >> s1 >> s2 >> w;        int id1 = transform (s1);        int id2 = transform (s2);        weight[id1] += w;        weight[id2] += w;        G[id1][id2] += w;        G[id2][id1] += w;    }    dfsTrave ();    cout << result.size () << endl;    for (map<string , int>::iterator it = result.begin (); it != result.end (); it++)        cout << it->first << " " << it->second << endl;    return 0;}