1114 Family Property （25 分）

1114 Family Property （25 分）

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 Mestate 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; Mestate 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 AVGsets AVGarea

where ID is the smallest ID in the family; M is the total number of family members; AVGsets is the average number of sets of their real estate; and AVGarea 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

输出就是每个集合中id最小的，总共多少套房子，平均每人的房子数量和平均每人的房子面积，

先按照平均每人的房子面积从大到小输出如果相等的话那么就按照id从小到大输出

思路：并查集的题，这题水就是处理有一点点烦

代码：

#include<bits/stdc++.h>
using namespace std;
struct node{int upset,squ,cnt;node(){cnt = 1;upset = 0;squ = 0;}
}s[10005];
struct GG{double cnt,upset,squ;int id;
};
int pre[10005];
bool flag[10005]; //刚开始不都是false  如果有出现就true
int find(int x){if( x == pre[x])return x;return pre[x] = find(pre[x]);
}
void merge(int x, int y){int fx = find(x);int fy = find(y);if(fx > fy)pre[fx] = fy;else if(fx < fy)pre[fy] = fx; return;
}
bool cmp(GG a, GG b){if(a.squ != b.squ)return a.squ > b.squ;else return a.id < b.id;
}
int main(){int n;scanf("%d", &n);for(int i = 1; i <= 10000; i ++)pre[i] = i;int me, fa, mo, cnt, child;for(int i = 1; i <= n; i ++){scanf("%d %d %d",&me,&fa,&mo);flag[me] = true; if(fa != -1){merge(me, fa);flag[fa] = true;}   if(mo != -1){merge(me, mo);flag[mo] = true;}  scanf("%d",&cnt);for(int j = 1; j <= cnt; j ++ ){scanf("%d",&child); merge(child, me);flag[child] = true;}  scanf("%d %d",&s[me].upset, &s[me].squ);}set<int>st;for(int i = 10000; i >= 0; i--)  //这边wa了第四个测试点因为0---10000  我写成1----10000{if(flag[i] == true){int x = find(i);//找到它的祖先st.insert(x);if(x != i){s[x].cnt += s[i].cnt;s[x].squ += s[i].squ;s[x].upset += s[i].upset; }}}set<int>::iterator it = st.begin();vector<GG>vec;while(it!=st.end()){GG gg;gg.id = *it;gg.cnt = s[*it].cnt;gg.squ =s[*it].squ * 1.0 / s[*it].cnt * 1.0;gg.upset = s[*it].upset * 1.0 / s[*it].cnt * 1.0;vec.push_back(gg);it++;}sort(vec.begin(),vec.end(),cmp);printf("%d\n",vec.size());for(int i = 0 ; i < vec.size(); i++)printf("%04d %.0lf %.3lf %.3lf\n",vec[i].id, vec[i].cnt,vec[i].upset, vec[i].squ);return 0;
}