[C++] PAT 1034 Head of a Gang (30分)

Sample Input1：

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 Output1：

2
AAA 3
GGG 3

Sample Input2：

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 Output2：

0

题解

代码参考了算法笔记
1.该题是无向图的DFS遍历，计算出连通分量
2.使用二维数组构建邻接矩阵，由于名字给的是字母，所以需要构造两个map，一个为map,一个为map,这样数字和名字可以一一对应
3.每次访问完一条边后，将该边weight=0，为的是去除无限递归

#include 
#include 
#include 
using namespace std;
const int MAXV = 2010;

int G[MAXV][MAXV]={0},weight[MAXV]={0};
bool vis[MAXV]={false};
map<string,int> ans;
map<string,int> string2Int;
map<int,string> int2String;
int n,k,numPerson=1;//n为通话数，k下界，numPerson人数

 




int change(string str){
	if(string2Int.find(str) != string2Int.end()){
		return string2Int[str];
	}else{
		string2Int[str] = numPerson;
		int2String[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;
			if(vis[i] == false){
				DFS(i,head,numMember,totalValue);
			} 
		}
	}
}

void DFSTravel(){
	for(int i = 0;i < numPerson;i++){
		if(vis[i] == false){
			int head = i,numMember = 0,totalValue=0;//head头目，numMember成员数，totalValue总价值
			
			DFS(i,head,numMember,totalValue);
			
			if(numMember > 2 && totalValue > k){
				ans[int2String[head]] = numMember;
			} 
		}
	}
}

int main(){
	int w;//临时权重
	string str1,str2;
	
	cin >> n >> k;
	
	for(int i = 0;i < n ;i++){
		cin >> str1 >> str2 >> w;
		
		int id1 = change(str1);
		int id2 = change(str2);
		
		weight[id1] += w;
		weight[id2] += w;
		
		G[id1][id2] += w;
		G[id2][id1] += w;
		
	}
	
	DFSTravel();
	
	cout<< ans.size() << endl;
	
	map<string,int>::iterator i;
	for(i = ans.begin();i != ans.end();i++){
		cout << i->first << " " << i->second << endl;
	}
	
	return 0;
}