PAT 1114. Family Property

最新推荐文章于 2020-09-04 17:24:55 发布

原创最新推荐文章于 2020-09-04 17:24:55 发布 · 503 阅读

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本文介绍了一个用于统计家庭成员间财产平均值的算法实现。通过构建人员之间的关系图，并使用深度优先搜索遍历整个图来确定每个家庭的成员及财产总数，最终计算出每个家庭的平均财产面积和数量。

思路是之前一直遇到的问题：联通分量的求解，

刚开始考虑用map吧string都转成int，以为这样可以减小运算量，但是这样就会导致一个问题：输入的顺序是乱的，可以某个ID要在后面出现，而前面就要用，所以就hash不到值

无奈，就用暴力的方法，ID不是4位数嘛，就构造一个10000*10000的bool数组，但是在累加资产的时候又遇到了问题，某个人也许没有资产在他的名下，但是他也可能与有名下资产的人有关系，因此还要嵌套一层循环。

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 ... Child_k 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; Child_i's are the ID's of his/her children; M_estateis 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

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdio>

using namespace std;

typedef struct person{
	int id, fid, mid, clen;
	vector<int> cs;
	//set<person> cs;
	int M, A;
};

typedef struct family{
	vector<int> vect;
	double sets, areas, avg_sets, avg_areas;
};

int N;
bool f[10000][10000];
person ps[1001];
bool marked[1001];
vector<family *> rst;
family *temp;
map<int, int> p;

bool cmp(family* f1, family* f2) {
	if(f1->avg_areas > f2->avg_areas)	return true;
	if(f1->avg_areas < f2->avg_areas)	return false;
	if(f1->vect[0] < f2->vect[0])	return true;
	if(f1->vect[0] > f2->vect[0])	return false;
}

void dfs(int start)
{
	int idx = ps[start].id;
	for(int i=0; i<10000; i++) {
		if(f[idx][i]) {
			if(find(temp->vect.begin(), temp->vect.end(), i) == temp->vect.end())
				temp->vect.push_back(i);

			for(int j=0; j<10000; j++) {
				if(j == i || f[i][j]) {
					if(p.find(j) != p.end() && !marked[p.find(j)->second]) {				
						int next = p.find(j)->second;
						marked[next] = true;
						temp->areas += ps[next].A;
						temp->sets += ps[next].M;
						dfs(next);
					}
				}
			}
		}
	}
}

int main()
{
	cin >> N;
	for(int i=0; i<N; i++) {
		cin >> ps[i].id >> ps[i].fid >> ps[i].mid >> ps[i].clen;
		for(int j=0; j<ps[i].clen; j++) {
			int c;	cin >> c;
			ps[i].cs.push_back(c);
			if(c != -1)	{f[ps[i].id][c] = true;	f[c][ps[i].id] = true;}
		}
		if(ps[i].fid!=-1)	{f[ps[i].id][ps[i].fid] = true;	f[ps[i].fid][ps[i].id] = true;}
		if(ps[i].mid!=-1)	{f[ps[i].id][ps[i].mid] = true;	f[ps[i].mid][ps[i].id] = true;}
		if(ps[i].fid!=-1 && ps[i].mid!=-1) {
			f[ps[i].fid][ps[i].mid] = true;	f[ps[i].mid][ps[i].fid] = true;
		}
		cin >> ps[i].M >> ps[i].A;
		p.insert(pair<int, int>(ps[i].id, i));
	}

	for(int i=0; i<N; i++) {
		if(!marked[i]) {
			temp = new family();
			temp->vect.push_back(ps[i].id);
			temp->sets = ps[i].M;
			temp->areas = ps[i].A;
			marked[i] = true;
			dfs(i);
			sort(temp->vect.begin(), temp->vect.end());
			int len = temp->vect.size();
			temp->avg_sets = temp->sets/len;
			temp->avg_areas = temp->areas/len;
			rst.push_back(temp);
		}
	}

	sort(rst.begin(), rst.end(), cmp);
	printf("%d\n", rst.size());
	for(int i=0; i<rst.size(); i++) {
		printf("%04d %d %.3lf %.3lf\n", rst[i]->vect[0], rst[i]->vect.size(), 
						rst[i]->avg_sets, rst[i]->avg_areas);
	}

	

	return 0;
}