pat 甲1114. Family Property (dfs)

1114. Family Property (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

This time, you are supposed to help us collect the data for family-owned property. Given each person's family members, and the estate（房产）info under his/her own name, we need to know the size of each family, and the average area and number of sets of their real
estate.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=1000). Then N lines follow, each gives the infomation of a person who owns estate in the format:

ID Father Mother k Child1 ... Childk M_estate Area

where ID is a unique 4-digit identification number for each person; Father and Mother are the ID's of this person's parents (if a parent has passed away, -1 will be given instead); k (0<=k<=5)
is the number of children of this person; Childi's are the ID's of his/her children; M_estate is the total number of sets of the real estate under his/her name; and Area is
the total area of his/her estate.

Output Specification:

For each case, first print in a line the number of families (all the people that are related directly or indirectly are considered in the same family). Then output the family info in the format:

ID M AVG_sets AVG_area

where ID is the smallest ID in the family; M is the total number of family members; AVG_sets is the average number of sets of their real estate; and AVG_area is the average area. The average
numbers must be accurate up to 3 decimal places. The families must be given in descending order of their average areas, and in ascending order of the ID's if there is a tie.
Sample Input:

10
6666 5551 5552 1 7777 1 100
1234 5678 9012 1 0002 2 300
8888 -1 -1 0 1 1000
2468 0001 0004 1 2222 1 500
7777 6666 -1 0 2 300
3721 -1 -1 1 2333 2 150
9012 -1 -1 3 1236 1235 1234 1 100
1235 5678 9012 0 1 50
2222 1236 2468 2 6661 6662 1 300
2333 -1 3721 3 6661 6662 6663 1 100

Sample Output:

3
8888 1 1.000 1000.000
0001 15 0.600 100.000
5551 4 0.750 100.000

tips:题目其实给出了家属关系和家族属性两组数据
用g来存储关系，结构体node存放属性.深搜顺便得到联通块的个数。

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

int n,sz;
int book[10005];
vector<int>g[10005];
struct node{
int set,area;
};

node a[10005];
int minid,num,ts,ta;
struct node1{
int minid,num;
double s,a;
friend bool operator<(node1 x,node1 y)
{
if(x.a!=y.a)return x.a>y.a;
else
return x.minid<y.minid;
}
};
node1 ans[10005];
void dfs(int x)
{
book[x]=1;

minid=min(x,minid);
num++;
ts+=a[x].set;
ta+=a[x].area;

for(int i=0;i<g[x].size();i++)
{
if(!book[g[x][i]])dfs(g[x][i]);
}
}
int main()
{
cin>>n;
memset(book,-1,sizeof(book));

for(int j=1;j<=n;j++)
{
int x,y,z;cin>>x>>y>>z;
book[x]=book[y]=book[z]=0;

if(y!=-1)g[x].push_back(y),g[y].push_back(x);
if(z!=-1)g[x].push_back(z),g[z].push_back(x);

int k;cin>>k;
for(int i=0;i<k;i++)
{
int t;cin>>t;book[t]=0;
g[x].push_back(t),g[t].push_back(x);
}

cin>>a[x].set>>a[x].area;
}

for(int i=0;i<=10000;i++)
{
minid=11100;num=ts=ta=0;
if(!book[i])
{
dfs(i);
ans[sz].minid=minid;
ans[sz].num=num;
ans[sz].s=ts*1.0/num;
ans[sz].a=ta*1.0/num;
sz++;
}

}
cout<<sz<<endl;
sort(ans,ans+sz);
for(int i=0;i<sz;i++)
printf("%04d %d %.3lf %.3lf\n",ans[i].minid,ans[i].num,ans[i].s,ans[i].a);
return 0;
}