PAT甲级_1034 Head of a Gang (30 分)

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1、1034 Head of a Gang (30 分)

题目大意：给出一张图，得出每个连通分量，其点中连接边的权和最大的点为head，点的数量大于k，则输出其head及点数；

将每个点的字符串与整型字符对应，在需要将字符串储存进结果map进行输出时再使用inttostr进行转化；DFS整张图，得到几个连通分量，在输入过程中顺便完成每个点权值的计算；在DFS过程中将遍历过的边重新赋值为0，以免在遍历到下一个点（或其他点）时重复计算；

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
int he[2001][2001], edge[2001][2001], weight[2001], k;
bool visit[2001];
map<string, int> strtoint;
map<int, string> inttostr;
map<string, int> final_;
int no=1;
int conser(string s) {
	if (strtoint[s]==0) {
		strtoint[s] = no;
		inttostr[no] = s;
		return no++;
	}
	else{
		return strtoint[s];
	}
}

void DFS(int u, int &head, int &number, int &totalweight) {
	visit[u] = true;
	number++;
	for (int i = 1; i < no; i++) {
		if (weight[head] < weight[u]) {
			head = u;
		}
	}
	for (int i = 1; i < no; i++) {
		if (edge[u][i] > 0) {
			totalweight += edge[u][i];
			edge[u][i] = edge[i][u] = 0;
			if (visit[i] == false) DFS(i, head, number, totalweight);
		}
	}
}

void trace() {
	for (int i = 1; i < no; i++) {
		if (visit[i] == false) {
			int head = i, number = 0, totalweight = 0;
			DFS(i, head, number, totalweight);
			if (number > 2 && totalweight > k) {
				final_[inttostr[head]] = number;
			}
		}
	}
}

int main() {
	int n, w;
	scanf("%d %d", &n, &k);
	string s1, s2;
	for (int i = 0; i < n; i++) {
		cin>>s1>>s2>>w;
		int e = conser(s1);
		int r = conser(s2);
		weight[e] += w;
		weight[r] += w;
		edge[e][r] += w;
		edge[r][e] += w;
	}
	trace();
	printf("%d\n", final_.size());
	for (auto it = final_.begin(); it != final_.end(); it++) {
		cout<<it->first<<" "<<it->second<<endl;
	}
	return 0;
}