PTA_PAT甲级_1034 Head of a Gang (30分)

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

题意：
给出若干人的通话记录，若有联系的某几个人（大于2个）的通话总时长大于一个阈值，则这几个人构成一个犯罪团伙，其中与其他人通话时间和最长的人为头目，输出犯罪团伙的个数和各个团伙的头目和团伙成员数

分析：
用map映射名字与编号，用两次DFS分别获取每个连通块内的头目和人数以及每个连通块的信息

代码：

#include
#include
#include
using namespace std;

#define MAXN 2010
map<int, string> intToString;//编号->姓名 
map<string, int> stringToInt;//姓名->编号 
map<string, int> Gang;//head->人数
int G[MAXN][MAXN]={0}, weight[MAXN]={0};//邻接矩阵，点权
int N, K, numPerson=0;//边数,下限,总人数
bool vis[MAXN] = {false};//标记是否被访问 

//DFS函数访问单个连通块，nowVisit为当前访问的编号
//head为头目，numMember为成员编号，totalValue为连通块的总边权 
void DFS(int nowVisit, int& head, int& numMember, int& totalValue){
	numMember++;//成员人数加1 
	vis[nowVisit] = true; 
	if(weight[nowVisit]>weight[head]) head = nowVisit;//当前节点点权更大则更新头目
	for(int i=0;i<numPerson;i++)//枚举所有人 
		if(G[nowVisit][i]>0){//如果从当前节点能到达 
			totalValue += G[nowVisit][i];//连通块总边权增加该边权
			G[nowVisit][i] = G[i][nowVisit] = 0;//**删除此边，防止回头**
			if(vis[i]==false) DFS(i, head, numMember, totalValue);//若i为被访问则递归访问i 
		} 
}

//DFSTrave函数遍历整个图，获取每个连通块的信息 
void DFSTrave(){
	for(int i=0;i<numPerson;i++)//枚举所有人 
		if(vis[i]==false){//如果i未被访问 
			int head=i, numMember=0, totalValue=0;//头目、成员数及总边权
			DFS(i, head, numMember, totalValue);//遍历i所在的连通块
			if(numMember>2 && totalValue>K)//如果成员数大于2且总边权大于K 
				Gang[intToString[head]] = numMember; //head的人数为numMember 
		}
}

//change函数返回姓名str对应的编号
int change(string str){
	if(stringToInt.find(str)!=stringToInt.end()) return stringToInt[str];//已出现过则返回编号
	else{
		stringToInt[str] = numPerson;//str的编号为numPerson
		intToString[numPerson] = str;//numPerson对应str
		return numPerson++;//总人数加1 
	} 
} 

int main()
{
    cin>>N>>K;
	for(int i=0;i<N;i++){
		string str1, str2;
		int w;
		cin>>str1>>str2>>w;//两个端点字符和点权 
		int id1 = change(str1);//转换为编号 
		int id2 = change(str2); 
		weight[id1] += w;//点权增加w 
		weight[id2] += w;
		G[id1][id2] += w;//边权增加w 
		G[id2][id1] += w; 
	}
	DFSTrave();//遍历整个图的所有连通块，获取Gang的信息
	cout<<Gang.size()<<endl;
	for(map<string,int>::iterator it = Gang.begin();it!=Gang.end();it++)
		cout<<it->first<<" "<<it->second<<endl; 
    return 0;
}