【学习笔记】〖九度OJ〗题目1446：Head of a Gang

题目1446：Head of a Gang

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：622

解决：144

题目描述：

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.

输入：

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.

输出：

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.

样例输入：

8 59AAA BBB 10BBB AAA 20AAA CCC 40DDD EEE 5EEE DDD 70FFF GGG 30GGG HHH 20HHH FFF 108 70AAA BBB 10BBB AAA 20AAA CCC 40DDD EEE 5EEE DDD 70FFF GGG 30GGG HHH 20HHH FFF 10

样例输出：

2AAA 3GGG 30

来源：

2012年浙江大学计算机及软件工程研究生机试真题

用了并查集，之后遍历统计各组信息，提交之后想到可以在合并集合时就把统计信息顺路做了，这样能少一次遍历，

时间紧迫就不改了。

题目说n小于等于1000，是电话的个数，而不是人的个数，所以并查集要开到2000，没好好读题在这错了测试点。。

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<vector>using namespace std;#define MAX 2003#define NL 6class People{public:    char name[NL];    int tTime;//总通话时间    int father;//用于并查集         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];//并查集 class Gang{public:    int head;//head在tree中的位置    int num;//人数    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];//团伙 int num;//当前并查集中元素个数int findPeople(char name[NL])//在并查集中找到name对应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;//并查集清空        cin >> k;        for (i=0; i<n; i++)        {            char name1[NL];            char name2[NL];            int time;            cin >> name1>> name2>> time;            //读入people信息            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;            }             //合并集合            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);            //第一个结点            if (g[root].head == -1)            {                g[root].head = i;            }            //最长时间为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;}