PAT.A1034 Head of a Gang-优快云博客

本文链接：https://blog.youkuaiyun.com/mokena1/article/details/83834356

本文介绍了一种警方通过分析电话记录来识别犯罪团伙及其头目的方法。通过对电话通话的时间长度进行加权，形成人际关系网络，并定义超过一定阈值的紧密联系群体为犯罪团伙。每个团伙中总权重最大的个体被视为头目。文章详细解释了输入输出规格及示例。

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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.

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 Nlines 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:

#include<iostream>
#include<queue>
#include<map>
#include<string>
using namespace std;
const int maxn = 1010;

int n, k,l,number=0,d1,d2,sum,p=0,q,maxd,maxnum;
int edge[maxn][maxn] = { 0 };
bool vist[maxn] = { false };
map<string,int> name,z;
map<int, string> name2;
int gang[maxn] = { 0 }, node[maxn] = { 0 };
string str1, str2;

void dfs(int u) {
	vist[u] = true;
	for (int i = 0; i < number; i++) {
		if (!vist[i] && edge[u][i] != 0) {
			//cout << name2[i] << endl;
			sum += node[i];
			maxd++;
			if (gang[i] >q) {
				q = gang[i];
				maxnum = i;
			}
			dfs(i);
		}
	}
}   
void dfstra() {
	for (int i = 0; i < number; i++) {
		sum = node[i];
		maxd = 1;
		q = gang[i];
		maxnum = i;
		if (!vist[i]) {
			dfs(i);
			//printf("\n\n");
			//cout << sum <<" "<<maxd<<endl;
			if (sum>k&&maxd > 2) {
			//	cout << sum <<" "<<maxd << endl;
				p++;
				z[name2[maxnum]] = maxd;
			}
		}
	}
}
void ini(string str,int l,int &d) {
	if (name.find(str) == name.end()) {
		name[str] = number;
		name2[number++] = str;
		}
		d = name[str];
		gang[d] += l;
}
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		cin >> str1 >> str2 >> l;
		ini(str1,l,d1);
		ini(str2,l,d2);
		node[d2]+= l;
		edge[d1][d2] = 1;
	}
	
	dfstra();
	cout << p << endl;
	if (p > 0) {
		map<string, int>::iterator it;
		for (it = z.begin(); it != z.end(); it++) {
			cout << it->first << " " << it->second << endl;
		}
	}

	
	
	return 0;
}