PAT (Advanced Level) Practise 1034 Head of a Gang (30)

1034. Head of a Gang (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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

题意：给你n对人的通话时间，gang 的定义是一群人，至少有 3 个人，这群人中每个人之间都通过通话相连，且整个群体的通话时长超过一个阈值。整个 gang 的团体中，拥有的电话时长最长的人就是头目，输出一共有几个gang，每个gang的头目和每个gang有多少人

解题思路：并查集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

map<string,int>mp;
int n,m,f[10009],sum[10009],sum1[10009],vis[10009];
char s1[10009][30],s2[10009][30],s3[10009][30];

struct node
{
char ch[30];
int sum;
friend bool operator<(node a,node b)
{
return strcmp(a.ch,b.ch)<0;
}
}ans[10009];

int Find(int x)
{
return f[x]==x?x:f[x]=Find(f[x]);
}

int main()
{
while(~scanf("%d%d",&n,&m))
{
int cnt=1,w,tot=0;
for(int i=1;i<=10000;i++) f[i]=i;
mp.clear();
memset(sum,0,sizeof sum);
memset(sum1,0,sizeof sum1);
for(int i=1;i<=n;i++)
{
scanf("%s%s%d",s1[i],s2[i],&w);
if(!mp[s1[i]]) mp[s1[i]]=cnt,strcpy(s3[cnt],s1[i]),cnt++;
if(!mp[s2[i]]) mp[s2[i]]=cnt,strcpy(s3[cnt],s2[i]),cnt++;
int x=mp[s1[i]],y=mp[s2[i]];
sum[x]+=w,sum[y]+=w;
}
for(int i=1;i<=n;i++)
{
int x=mp[s1[i]],y=mp[s2[i]];
int xx=Find(x),yy=Find(y);
if(xx!=yy&&sum[xx]>sum[yy]) f[yy]=xx;
else if(xx!=yy) f[xx]=yy;
}
memset(vis,0,sizeof vis);
for(int i=1;i<cnt;i++)
{
vis[Find(i)]++;
sum1[Find(i)]+=sum[i];
}
for(int i=1;i<cnt;i++)
{
if(Find(i)==i&&vis[i]>2&&sum1[i]/2>m)
strcpy(ans[tot].ch,s3[i]),ans[tot++].sum=vis[i];
}
printf("%d\n",tot);
sort(ans,ans+tot);
for(int i=0;i<tot;i++)
printf("%s %d\n",ans[i].ch,ans[i].sum);
}
return 0;
}