PAT A1034 Head of a Gang

• 通话记录有1000条，人数可能为2000
• 给的人名是string，用两个map记录数字转换
• 每遍历过一条边后就把这条边的权值设为0，防止出现回路遍历死循环

#include <cstdio>
#include <vector>
#include <iostream>
#include <queue>
#include <algorithm>
#include <map>

using namespace std;

const int maxn = 2010;
map<string, int> stringToint;//姓名-编号
map<int, string> intTostring;//编号-姓名
map<string, int> ans;
int G[maxn][maxn] = { 0 }, weight[maxn] = { 0 };
int n,k,numPerson=0;
bool vis[maxn] = { false };

int change(string str)
{
	if (stringToint.find(str) != stringToint.end()) {
		return stringToint[str];
	}
	else {
		stringToint[str] = numPerson;
		intTostring[numPerson] = str;
		return numPerson++;
	}
}

void DFS(int nowVisit, int& head, int& numMember, int& totalValue) {
	numMember++;
	vis[nowVisit] = true;
	if (weight[nowVisit] > weight[head]) {
		head = nowVisit;
	}
	for (int i = 0; i < numPerson; i++) {
		if (G[nowVisit][i] != 0) {
			totalValue += G[nowVisit][i];
			G[nowVisit][i] = G[i][nowVisit] = 0;//删除边，防止回头
			if (!vis[i]) {
				DFS(i, head, numMember, totalValue);
			}
		}
	}
}

void DFSTrave() {
	for (int i = 0; i < numPerson; i++) {
		if (vis[i] == false) {
			int head = i, numMember = 0,totalValue=0;
			DFS(i, head, numMember, totalValue);
			if (numMember > 2 && totalValue > k) {
				ans[intTostring[head]] = numMember;
			}
		}

	}
}
int main()
{
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		string t1, t2;
		int w;
		cin >> t1 >> t2 >> w;
		int id1 = change(t1);
		int id2 = change(t2);
		weight[id1] += w;
		weight[id2] += w;
		G[id1][id2] += w;
		G[id2][id1] += w;

	}
	DFSTrave();
	cout << ans.size() << endl;
	map<string, int>::iterator it;
	for (it = ans.begin(); it != ans.end(); it++) {
		cout << it->first << " " << it->second << endl;
	}
	return 0;
}

标签：Head,PAT,int,head,numPerson,numMember,A1034,nowVisit,include
来源： https://blog.csdn.net/Mind_The_Gap/article/details/114486490