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.

8 59AAA BBB 10BBB AAA 20AAA CCC 40DDD EEE 5EEE DDD 70FFF GGG 30GGG HHH 20HHH FFF 10Sample Output 1:
2AAA 3GGG 3Sample Input 2:
8 70AAA BBB 10BBB AAA 20AAA CCC 40DDD EEE 5EEE DDD 70FFF GGG 30GGG HHH 20HHH FFF 10Sample Output 2:
0
給出1000條以內的通話記錄A B和權值w，和閾值k，如果一個團伙人數超過2人并且通話總權值超過k，
令團伙里面的自身權值的最大值為頭目，輸出所有滿足條件的團伙的頭目，和他們團伙里面的人數
#include <iostream>#include <map>using namespace std;map<string, int> stringToInt;map<int, string> intToString;map<string, int> ans;int idNumber = 1, k;//總共的人數idNumberint stoifunc(string s) {    if(stringToInt[s] == 0) {        stringToInt[s] = idNumber;        intToString[idNumber] = s;        return idNumber++;    } else {        return stringToInt[s];    }}//把string轉化成數字，比如AAA->0,BBB->1......但未必是按字母表順序int G[2010][2010], weight[2010];//G[][]存儲每條邊的權值，weight[]存每個點的權值。bool vis[2010];void dfs(int u, int &head, int &numMember, int &totalweight) {    vis[u] = true;//numMember小組成員數,u當前節點，head頭結點，totalweight總權值    numMember++;    if(weight[u] > weight[head])        head = u;    for(int v = 1; v < idNumber; v++) {        if(G[u][v] > 0) {            totalweight += G[u][v];            G[u][v] = G[v][u] = 0;//加過了標記上不再加了            if(vis[v] == false)                dfs(v, head, numMember, totalweight);        }    }}void dfsTrave() {    for(int i = 1; i < idNumber; i++) {        if(vis[i] == false) {            int head = i, numMember = 0, totalweight = 0;            dfs(i, head, numMember, totalweight);            if(numMember > 2 && totalweight > k)                ans[intToString[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 = stoifunc(s1);        int id2 = stoifunc(s2);        weight[id1] += w;        weight[id2] += w;        G[id1][id2] += w;        G[id2][id1] += w;    }    dfsTrave();    cout << ans.size() << endl;    for(map<string, int>::iterator it = ans.begin(); it != ans.end(); it++)        cout << it->first << " " << it->second << endl;    return 0;}