PAT 1114 Family Property[并查集][难]

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 Nlines 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

题目大意：找出一个家庭中人均有几套房产，和人均面积；输入中为当前人的ID 父亲ID 母亲ID 当前人拥有几套当前人拥有总面积。如果父母去世了，则用-1表示。输出要求按平均房产套数递减，如果相同，那么按ID递增排列。

//确实是使用并查集。

//遇到问题：遇到像8888这种单户的，该怎么处理，我写的里边似乎处理不了。

#include <iostream>

#include <algorithm>

#include <vector>

#include <map>

#include<cstdio>

using namespace std;

struct Peo{

    int father,estate,area;

    Peo(){father=-;}

}peo[];

map<int,int> mp;

struct Far{

    int id,mem;

    double avge,avga;

};

int findF(int a){

    if(peo[a].father==-)return a;

    int k=peo[a].father;

    while(peo[k].father!=-){

        peo[a].father=peo[k].father;

        a=k;

        k=peo[a].father;

        //cout<<"wwww";

    }

    //cout<<"fff";

    return k;

}

void unionF(int a,int b){

    int fa=findF(a);

    int fb=findF(b);

//    if(fa>fb)peo[fa].father=fb;

//    else peo[fb].father=fa;//这里出现了问题啊，得判断是否等于，等于的话，啥也不用了。

    if(fa>fb)peo[fa].father=fb;

    else if(fa<fb) peo[fb].father=fa;

    //cout<<"uuuu";

}

bool cmp(Far & a,Far & b){

    return a.avge!=b.avge?a.id<b.id:a.avge>b.avge;

}

int main() {

    int n;

    cin>>n;

    //fill(father,father+10000,-1);

    int id,fa,mo,k,child;

    for(int i=;i<n;i++){

        cin>>id>>fa>>mo>>k;

        //先将当前的人和父亲母亲合并

        if(fa!=-)

            unionF(id,fa);

        if(mo!=-)

            unionF(id,mo);

        for(int j=;j<k;j++){

            cin>>child;

            unionF(id,child);

        }

       // peo[id].father=id;

        cin>>peo[id].estate>>peo[id].area;

    }

    vector<int> vt[];

    for(int i=;i<;i++){

        if(peo[i].father!=-){

            //cout<<"hh";

            mp[peo[i].father]++;

            //怎么记录每个簇里有谁呢？

            vt[peo[i].father].push_back(i);

        }

    }

    //最终还得sort一下。

    cout<<mp.size()<<'\n';

    vector<Far> far;

    for(auto it=mp.begin();it!=mp.end();it++){

        int id=it->first;

        int mem=vt[id].size();

        int tote,tota;

        tote=peo[id].estate;

        tota=peo[id].area;

        for(int i=;i<vt[id].size();i++){

            tote+=peo[vt[id][i]].estate;

            tota+=peo[vt[id][i]].area;

        }

        far.push_back(Far{id,mem,tote*1.0/mem,tota*1.0/mem});

    }

    sort(far.begin(),far.end(),cmp);

    for(int i=;i<far.size();i++){

        printf("%04d %d %.3f %.3f\n",far[i].id,far[i].mem,far[i].avge,far[i].avga);

    }

    return ;

}

//写成了这样，最终决定放弃！

代码转自：https://www.liuchuo.net/archives/2201

#include <cstdio>

#include <algorithm>

using namespace std;

struct DATA {

    int id, fid, mid, num, area;

    int cid[];//最多有10个孩子。

}data[];

struct node {

    int id, people;

    double num, area;

    bool flag = false;

}ans[];

int father[];

bool visit[];

int find(int x) {

    while(x != father[x])//这样真的好简便，我为啥写那么复杂呢。

        x = father[x];

    return x;

}

void Union(int a, int b) {

    int faA = find(a);

    int faB = find(b);

    if(faA > faB)

        father[faA] = faB;

    else if(faA < faB)

        father[faB] = faA;

}

int cmp1(node a, node b) {

    if(a.area != b.area)

        return a.area > b.area;

    else

        return a.id < b.id;

}

int main() {

    int n, k, cnt = ;

    scanf("%d", &n);

    for(int i = ; i < ; i++)

        father[i] = i;//将父亲设置为自己。

    for(int i = ; i < n; i++) {

        scanf("%d %d %d %d", &data[i].id, &data[i].fid, &data[i].mid, &k);

        //直接读进来，不使用中间变量。

        //并没有使用下标作为id啊。

        visit[data[i].id] = true;//标记出现过了。

        if(data[i].fid != -) {

            visit[data[i].fid] = true;

            Union(data[i].fid, data[i].id);//将id作为并查集中的关键字合并。

        }

        if(data[i].mid != -) {

            visit[data[i].mid] = true;

            Union(data[i].mid, data[i].id);

        }

        for(int j = ; j < k; j++) {

            scanf("%d", &data[i].cid[j]);

            visit[data[i].cid[j]] = true;

            Union(data[i].cid[j], data[i].id);

        }

        scanf("%d %d", &data[i].num, &data[i].area);

    }

    for(int i = ; i < n; i++) {

        int id = find(data[i].id);//找到当前人的父亲，

        ans[id].id = id;//现在这个ans中使用id作为下标索引了！

        ans[id].num += data[i].num;

        ans[id].area += data[i].area;

        ans[id].flag = true;

    }

    for(int i = ; i < ; i++) {

        if(visit[i])//如果它出现过。

       i     ans[find(i)].people++;

        if(ans[i].flag)//标记有几簇人家。

            cnt++;

    }

    for(int i = ; i < ; i++) {

        if(ans[i].flag) {

            ans[i].num = (double)(ans[i].num * 1.0 / ans[i].people);

            ans[i].area = (double)(ans[i].area * 1.0 / ans[i].people);

            //并没有多个num属性，都是用一个存的，一开始就设为double。

            //我居然还分开定义了。

        }

    }

    sort(ans, ans + , cmp1);

    printf("%d\n", cnt);

    for(int i = ; i < cnt; i++)

        printf("%04d %d %.3f %.3f\n", ans[i].id, ans[i].people, ans[i].num, ans[i].area);

    return ;

}

1.将Father初始化为了自己，那么find函数就简单了，学习了

2.使用一个bool数组来标记出现过的点和没出现过的点，就解决了一家只有一口的情况。

3.在求ans向量的时候，仍使用了find，其实不用map的，find找到父亲都是可以用的。

4.因为data[i].id里存的是已经出现过的id,对那些没出现过的ans.flag肯定不会被标记为true的！

这是错误理解：ans里存储的所有的10000个人的情况，对于那些没有出现过的id，ans[i].flag也是true，只不过它所有的数据都是0，在经过排序之后，再通过簇进行控制输出，其他的不输出。

总之，学习了。