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PAT (Advanced Level) 1114. Family Property (25)

时间:2016-07-06 00:05:13      阅读:180      评论:0      收藏:0      [点我收藏+]

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简单DFS。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

struct X
{
    int id;
    int Father,Mother;
    int k;
    int son[10];
    int M_estate;
    int Area;
}s[10000+10];
int n;
int f[10000+10];
struct Ans
{
    int id;
    int M;
    int sum_set;
    int sum_area;
}ans[10000+10];
int sz;
int indx[10000+10];

vector<int>g[10000+10];
int min_id;
int sum;
int sum_set;
int sum_area;

bool cmp(const Ans&a,const Ans&b)
{
    if(1.0*a.sum_area/a.M==1.0*b.sum_area/b.M) return a.id<b.id;
    return 1.0*a.sum_area/a.M>1.0*b.sum_area/b.M;
}

void dfs(int x)
{
    ans[sz].id=min_id;
    ans[sz].M=sum;
    ans[sz].sum_area=sum_area;
    ans[sz].sum_set=sum_set;

    for(int i=0;i<g[x].size();i++)
    {
        if(f[g[x][i]]==1) continue;
        f[g[x][i]]=1;

        min_id=min(min_id,g[x][i]);
        sum=sum+1;
        sum_set=sum_set+s[g[x][i]].M_estate;
        sum_area=sum_area+s[g[x][i]].Area;

        dfs(g[x][i]);
    }
}

int main()
{
    for(int i=0;i<=10000;i++)
    {
        s[i].Father=-1;
        s[i].Mother=-1;
        s[i].k=0;
        s[i].Area=0;
        s[i].M_estate=0;
    }

    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int ID; scanf("%d",&ID); indx[i]=ID;
        s[ID].id=ID;
        scanf("%d%d",&s[ID].Father,&s[ID].Mother);
        scanf("%d",&s[ID].k);
        for(int j=0;j<s[ID].k;j++) scanf("%d",&s[ID].son[j]);
        scanf("%d",&s[ID].M_estate);
        scanf("%d",&s[ID].Area);

        if(s[ID].Father!=-1){
            g[s[ID].id].push_back(s[ID].Father);
            g[s[ID].Father].push_back(s[ID].id);
        }
        if(s[ID].Mother!=-1){
            g[s[ID].id].push_back(s[ID].Mother);
            g[s[ID].Mother].push_back(s[ID].id);
        }
        for(int j=0;j<s[ID].k;j++)
        {
            g[s[ID].id].push_back(s[ID].son[j]);
            g[s[ID].son[j]].push_back(s[ID].id);
        }
    }

    memset(f,0,sizeof f); sz=0;
    for(int i=1;i<=n;i++)
    {
        if(f[s[indx[i]].id]==1) continue;
        min_id=s[indx[i]].id;
        sum=1;
        sum_set=s[indx[i]].M_estate;
        sum_area=s[indx[i]].Area;
        f[s[indx[i]].id]=1;
        dfs(s[indx[i]].id);
        sz++;
    }

    sort(ans,ans+sz,cmp);
    printf("%d\n",sz);
    for(int i=0;i<sz;i++)
    {
        printf("%04d %d %.3lf %.3lf\n",ans[i].id,ans[i].M,1.0*ans[i].sum_set/ans[i].M,1.0*ans[i].sum_area/ans[i].M);
    }

    return 0;
}

 

PAT (Advanced Level) 1114. Family Property (25)

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原文地址:http://www.cnblogs.com/zufezzt/p/5645230.html

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