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题意给出若干人之间的通话长度(视为无向边),这些通话将他们分为若干组。每个组的总边权设为该组内的所有通话的长度之和,而每个人的点权设为该人参与的通话长度之和。现在给定一个阙值K,且只要一个组的总边权

承蒙不弃 2022-02-27 阅读 20

#include<iostream>
#include<string>
#include<map>
using namespace std;
const int maxn=2010;
const int INF=1000000000;
map<int,string>intToString;
map<string,int>stringToInt;
map<string,int>Gang;
int G[maxn][maxn]={0},weight[maxn]={0};
int n,k,numPerson=0;
bool visit[maxn]={false};
void DFS(int nowVisit,int& head,int& numMember,int& totalValue){
    numMember++;
    visit[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(visit[i]==false){
                DFS(i,head,numMember,totalValue);
            }
        }
    }
}
void DFSTrave(){
    for(int i=0;i<numPerson;i++){
        if(visit[i]==false){
            int head=i,numMember=0,totalValue=0;
            DFS(i,head,numMember,totalValue);
            if(numMember>2&&totalValue>k){
                Gang[intToString[head]]=numMember;
            }
        }
    }
}
int change(string str){
    if(stringToInt.find(str)!=stringToInt.end()){
        return stringToInt[str];
    }else{
        stringToInt[str]=numPerson;
        intToString[numPerson]=str;
        return numPerson++;
    }
}
int main(){
    int w;
    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;
    }
    DFSTrave();
    cout<<Gang.size()<<endl;
    map<string,int>::iterator it;
    for(it=Gang.begin();it!=Gang.end();it++){
        cout<<it->first<<" "<<it->second<<endl;
    }
    return 0;
}

 

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