【PAT】1034. Head of a Gang (30)
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 threshold 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 10Sample Output 1:
2 AAA 3 GGG 3Sample 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 10Sample Output 2:
0
分析：题目写的不够清楚。
什么是gang？
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 threshold K.
（1）成员数据大于2 （2）相互间的通话时间总和大于K，例如：
AAA BBB 10 BBB AAA 20 AAA CCC 40则她们相互间的通话时间为10 + 20 + 40 。
以下代码有两组数据错误，得分为22，待改进：
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
#define INF 9999
int N,K,index;
vector<vector<int> > cost;
vector<bool> used;
vector<int> weight;
map<string,int> name2id;
map<int, string> id2name;
pair<int,int> bfs(int s){
queue<int> que;
que.push(s);
int total=0, max=weight[s], maxIndex=s, cnt=0;
while(!que.empty()){
int t = que.front();
que.pop();
cnt++;
total += weight[t];
if(weight[t]>max){
max = weight[t];
maxIndex = t;
}
used[t] = true;
for(int i=0; i<index; i++){
if( (!used[i])&&(cost[t][i]!=INF) ){
used[i] = true;
que.push(i);
}
}
}
if(total>K*2 && cnt>2)
return make_pair(maxIndex,cnt);
else
return make_pair(-1,-1);
}
int main(){
scanf("%d%d",&N,&K);
int i,time;
string stra, strb;
int a, b;
cost.resize(2*N,vector<int>(2*N,INF));
weight.resize(2*N,0);
index=0;
for(i=0; i<N; i++){
cin>>stra>>strb>>time;
if(name2id.find(stra)!=name2id.end()){
a = name2id[stra];
}else{
a = index;
name2id.insert(make_pair(stra,index));
id2name.insert(make_pair(index,stra));
index++;
}
if(name2id.find(strb)!=name2id.end()){
b = name2id[strb];
}else{
b = index;
name2id.insert(make_pair(strb,index));
id2name.insert(make_pair(index,strb));
index++;
}
if(cost[a][b]==INF){
cost[a][b] = 0;
cost[b][a] = 0;
}
cost[a][b] += time;
cost[b][a] += time;
weight[a] += time;
weight[b] += time;
}
used.resize(index-1,false);
vector<pair<int,int> > result;
int max = 0;
for(i=0; i<index; i++){
if(!used[i]){
pair<int, int> p;
p = bfs(i);
if(p.first==-1){
continue;
}
if(p.second==max){
result.push_back(p);
}else if(p.second>max){
result.clear();
result.push_back(p);
max = p.second;
}
}
}
printf("%d\n",result.size());
for(i=0; i<result.size(); i++){
cout<<id2name[result[i].first]<<" "<<result[i].second<<endl;
}
return 0;
}