Family Property
2016-12-04 21:08
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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:
Sample Output:
本来用并交集来写的,结果好复杂,还有bug,去网上搜了结果用 dfs 简单多了。。。
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define maxn 10000
struct node{
int id;
int num;
double est;
double area;
node(int id=0, int num=0,double est=0,double area=0):id(id),num(num),est(est),area(area){}
}p[maxn];
int e[maxn],ar[maxn],f[maxn];
int id,num;
double est,area;
vector<int> t[maxn];
bool cmp(node a,node b){
if(a.area!=b.area)
return a.area>b.area;
return a.id<b.id;
}
void dfs(int x)
{
id=min(id,x);
f[x]=0;
num++;
est+=e[x];
area+=ar[x];
for(int i=0;i<t[x].size();i++){
if(f[t[x][i]]==1)
dfs(t[x][i]);
}
}
int main() {
int a;
cin >> a;
int total=0;
for(int i=0;i<a;i++){
int x,y,z;
scanf("%d %d %d",&x,&y,&z);
f[x]=1;
if(y!=-1){
f[y]=1;
t[x].push_back(y);
t[y].push_back(x);
}
if(z!=-1){
f[z]=1;
t[x].push_back(z);
t[z].push_back(x);
}
int b;
scanf("%d",&b);
for(int j=0;j<b;j++){
scanf("%d",&z);
f[z]=1;
t[x].push_back(z);
t[z].push_back(x);
}
int c,d;
scanf("%d %d",&c,&d);
e[x]=c;
ar[x]=d;
}
for(int i=0;i<10000;i++){
if(f[i]!=1)
continue;
id=10000;num=0;est=0;area=0;
dfs(i);
p[total++]=node(id,num,est/(double)num,area/(double)num);
}
sort(p,p+total,cmp);
printf("%d\n",total);
for (int i = 0; i < total; i++)
{
printf("%04d %d %.3lf %.3lf\n", p[i].id, p[i].num, p[i].est, p[i].area);
}
return 0;
}
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 0.600 100.0005551 4 0.750 100.000
本来用并交集来写的,结果好复杂,还有bug,去网上搜了结果用 dfs 简单多了。。。
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define maxn 10000
struct node{
int id;
int num;
double est;
double area;
node(int id=0, int num=0,double est=0,double area=0):id(id),num(num),est(est),area(area){}
}p[maxn];
int e[maxn],ar[maxn],f[maxn];
int id,num;
double est,area;
vector<int> t[maxn];
bool cmp(node a,node b){
if(a.area!=b.area)
return a.area>b.area;
return a.id<b.id;
}
void dfs(int x)
{
id=min(id,x);
f[x]=0;
num++;
est+=e[x];
area+=ar[x];
for(int i=0;i<t[x].size();i++){
if(f[t[x][i]]==1)
dfs(t[x][i]);
}
}
int main() {
int a;
cin >> a;
int total=0;
for(int i=0;i<a;i++){
int x,y,z;
scanf("%d %d %d",&x,&y,&z);
f[x]=1;
if(y!=-1){
f[y]=1;
t[x].push_back(y);
t[y].push_back(x);
}
if(z!=-1){
f[z]=1;
t[x].push_back(z);
t[z].push_back(x);
}
int b;
scanf("%d",&b);
for(int j=0;j<b;j++){
scanf("%d",&z);
f[z]=1;
t[x].push_back(z);
t[z].push_back(x);
}
int c,d;
scanf("%d %d",&c,&d);
e[x]=c;
ar[x]=d;
}
for(int i=0;i<10000;i++){
if(f[i]!=1)
continue;
id=10000;num=0;est=0;area=0;
dfs(i);
p[total++]=node(id,num,est/(double)num,area/(double)num);
}
sort(p,p+total,cmp);
printf("%d\n",total);
for (int i = 0; i < total; i++)
{
printf("%04d %d %.3lf %.3lf\n", p[i].id, p[i].num, p[i].est, p[i].area);
}
return 0;
}
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