1034 Head of a Gang （边不重复 dfs+map）

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 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

　　

此题非常综合，值得研究，包括全局变量的设置（G[][],vis[]是dfs通用的），change和map的搭配，注意观察题目是节点不能重复遍历还是边不能重复遍历

#include<iostream>
#include<string>
#include<map>
using namespace std;

map<string,int> si;
map<int,string> is;
map<string,int> Gang;//head->人数
const int maxn=2020;
int G[maxn][maxn]={0},weight[maxn]={0};  
int n,m,numPerson=0;
bool vis[maxn]={0};

void dfs(int nowV,int &head,int &numMember,int &totalValue){
    vis[nowV]=1;
    numMember++;
    if(weight[nowV]>weight[head])
        head=nowV;
    for(int i=0;i<numPerson;i++){
        if(G[nowV][i]>0){//这道题边不能重复，不同于常规的节点遍历不重复 
            totalValue+=G[nowV][i];
            G[nowV][i]=G[i][nowV]=0;
            if(vis[i]==0){//递归访问 
                dfs(i,head,numMember,totalValue);
            }
        }
        
    }
}

void dfsTravel(){
    for(int i=0;i<numPerson;i++){
        if(vis[i]==0){
            int head=i,num=0,tot=0;
            dfs(i,head,num,tot);
            if(num>2&&tot>m)
                Gang[is[head]]=num;
        }
    }
}

int change(string str){//第一次可能存在需要赋值 
    if(si.find(str)!=si.end()) return si[str];
    else{
        si[str]=numPerson;
        is[numPerson]=str;
        return numPerson++;
    }
}

int main(){
    int c;
    cin>>n>>m;
    string a,b;
    for(int i=1;i<=n;i++){
        cin>>a>>b>>c;
        int id1=change(a);
        int id2=change(b);
        weight[id1]+=c;
        weight[id2]+=c;
        G[id1][id2]+=c;
        G[id2][id1]+=c;
    }
    dfsTravel(); 
    cout<<Gang.size()<<endl;
    map<string,int>::iterator it;//map自动按键从小到大排序 
    for(it=Gang.begin();it!=Gang.end();it++){
        cout<<it->first<<" "<<it->second<<endl;
    }
    return 0;
}