時間限制：1 秒

記憶體限制：128 兆

特殊判題：否

提交：622

解決：144

題目描述：

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 threthold 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.

輸入：

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.

輸出：

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.

樣例輸入：

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
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

樣例輸出：

2
AAA 3
GGG 3
0

時間緊迫就不改了。

題目說n小於等於1000，是電話的個數，而不是人的個數，所以並查集要開到2000，沒好好讀題在這錯了測試點。。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
#define MAX 2003
#define NL 6
class People
{
public:
    char name[NL];
    int tTime;//總通話時間
    int father;//用於並查集
     
    void init(char n[NL], int time, int f)
    {
        for (int i=0; i<NL; i++)
        {
            name[i] = n[i];
        }
        tTime = time;
        father = f;
    }
};
 
 
People tree[MAX];//並查集
 
class Gang
{
public:
    int head;//head在tree中的位置
    int num;//人數
    int time;
 
    Gang()
    {
        head = -1;
        num = 0;
        time = 0;
    }
    void clear()
    {
        head = -1;
        num = 0;
        time = 0;
    }
    bool operator < (Gang g) const
    {
        if (head == -1)
        {
            return false;
        }
 
        return strcmp(tree[head].name, tree[g.head].name) < 0;
    }
};
 
Gang g[MAX];//團伙
 
int num;//當前並查集中元素個數
int findPeople(char name[NL])//在並查集中找到name對應people的位置
{
    for (int i=0; i<num; i++)
    {
        if (strcmp(name,tree[i].name) == 0)
        {
            return i;
        }
    }
    return -1;
}
 
int findRoot(int t)
{
    if (tree[t].father == -1)
    {
        return t;
    }
 
    int tmp = findRoot(tree[t].father);
    tree[t].father = tmp;
    return tmp;
}
 
int main()
{
    int n, k, i;
    while (cin >> n)
    {
        num = 0;//並查集清空
        cin >> k;
        for (i=0; i<n; i++)
        {
            char name1[NL];
            char name2[NL];
            int time;
            cin >> name1>> name2>> time;
            //讀入people資訊
            int n1 = findPeople(name1);
            int n2 = findPeople(name2);
            if (n1 == -1)
            {
                tree[num].init(name1, time, -1);
                n1 = num;//重新定位people在tree中位置
                num++;
                 
            }
            else
            {
                tree[n1].tTime += time;
            }
 
            if (n2 == -1)
            {
                tree[num].init(name2, time, -1);
                n2 = num;
                num++;
            }
            else
            {
                tree[n2].tTime += time;
            }
 
            //合併集合
            n1 = findRoot(n1);
            n2 = findRoot(n2);
            if (n1 != n2)
            {
                tree[n1].father = n2;
            }
        }//end of for
 
 
        for (i=0; i<num; i++)
        {
            g[i].clear();
        }
 
        for (i=0; i<num; i++)
        {
            int root = findRoot(i);
            //第一個結點
            if (g[root].head == -1)
            {
                g[root].head = i;
            }
            //最長時間為head
            if (tree[g[root].head].tTime < tree[i].tTime)
            {
                g[root].head = i;
            }
            g[root].time += tree[i].tTime;
            g[root].num++;
        }
 
        vector<Gang> v;
        for (i=0; i<num; i++)
        {
            if (tree[i].father == -1 
                && g[i].num > 2 
                && g[i].time > 2*k)
            {
                v.push_back(g[i]);
            }
        }
 
        if (!v.empty())
        {
            sort(v.begin(), v.end());
            cout <<v.size() << endl;
            for (i=0; i<v.size(); i++)
            {
                cout << tree[v[i].head].name << " " << v[i].num<< endl;
            }
        }
        else
        {
            cout << 0 << endl;
        }
 
    }//end of while
 
    return 0;
}