1034. Head of a Gang (30)
2016-07-09 13:36
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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:
Sample Output 1:
Sample Input 2:
Sample Output 2:
一个团伙定义为人数多于两个人和他们之间总的通话时间超过给定的threshold。
1.输入每对人之间的通话时间,更新每个人的通话时间,且用并查集(因为名字是字符串,所以用map来查找),把有通话的两个人相连起来,分成几个集合(用set储存集合);
2.对于每一个集合,如果人数不超过2个就忽略;然后算出该集合总的通话时间(总通话时间等于集合中全部人的通话时间之和除以2)和求出通话时间最大的那个人(头目),如果不超过threshold忽略之;如果超过了则是一个团伙,用map储存头目和团伙人数;
3.输出结果。
代码:
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
map<string,string>Map;
string find(string n)
{
while(Map.find(n)!=Map.end()) n=Map
;
return n;
}
int main()
{
int n,k;
scanf("%d %d",&n,&k);
map<string,int>weight;
map<string,set<string> >gang;
for(int i=0;i<n;i++)
{
string a,b;
int t;
cin>>a>>b>>t;
weight[a]+=t;
weight[b]+=t;
string aa=find(a);
string bb=find(b);
if(aa!=bb)
{
Map[bb]=aa;
}
}
map<string,string>::iterator it0;
for(it0=Map.begin();it0!=Map.end();it0++)
{
string root=find(it0->first);
gang[root].insert(it0->first);
}
map<string,int>res;
map<string,set<string> >::iterator it;
set<string>::iterator sit;
for(it=gang.begin();it!=gang.end();it++)
{
int w=0;
set<string>tmp=it->second;
tmp.insert(it->first);
int m=tmp.size();
if(m<3) continue;
string idx;int Max=-1;
for(sit=tmp.begin();sit!=tmp.end();sit++)
{
w+=weight[*sit];
if(Max<weight[*sit])
{
Max=weight[*sit];
idx=*sit;
}
}
w/=2;
if(w>k)
{
res[idx]=m;
}
}
printf("%d\n",res.size());
map<string,int>::iterator mit;
for(mit=res.begin();mit!=res.end();mit++)
{
cout<<mit->first<<" "<<mit->second<<endl;
}
}
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 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
一个团伙定义为人数多于两个人和他们之间总的通话时间超过给定的threshold。
1.输入每对人之间的通话时间,更新每个人的通话时间,且用并查集(因为名字是字符串,所以用map来查找),把有通话的两个人相连起来,分成几个集合(用set储存集合);
2.对于每一个集合,如果人数不超过2个就忽略;然后算出该集合总的通话时间(总通话时间等于集合中全部人的通话时间之和除以2)和求出通话时间最大的那个人(头目),如果不超过threshold忽略之;如果超过了则是一个团伙,用map储存头目和团伙人数;
3.输出结果。
代码:
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
map<string,string>Map;
string find(string n)
{
while(Map.find(n)!=Map.end()) n=Map
;
return n;
}
int main()
{
int n,k;
scanf("%d %d",&n,&k);
map<string,int>weight;
map<string,set<string> >gang;
for(int i=0;i<n;i++)
{
string a,b;
int t;
cin>>a>>b>>t;
weight[a]+=t;
weight[b]+=t;
string aa=find(a);
string bb=find(b);
if(aa!=bb)
{
Map[bb]=aa;
}
}
map<string,string>::iterator it0;
for(it0=Map.begin();it0!=Map.end();it0++)
{
string root=find(it0->first);
gang[root].insert(it0->first);
}
map<string,int>res;
map<string,set<string> >::iterator it;
set<string>::iterator sit;
for(it=gang.begin();it!=gang.end();it++)
{
int w=0;
set<string>tmp=it->second;
tmp.insert(it->first);
int m=tmp.size();
if(m<3) continue;
string idx;int Max=-1;
for(sit=tmp.begin();sit!=tmp.end();sit++)
{
w+=weight[*sit];
if(Max<weight[*sit])
{
Max=weight[*sit];
idx=*sit;
}
}
w/=2;
if(w>k)
{
res[idx]=m;
}
}
printf("%d\n",res.size());
map<string,int>::iterator mit;
for(mit=res.begin();mit!=res.end();mit++)
{
cout<<mit->first<<" "<<mit->second<<endl;
}
}
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