第十章并查集 2 AcWing 1604. 家产

ゞ浴缸里的玫瑰 2024-03-31 12:17 79阅读 0赞

## 第十章 并查集 2 AcWing 1604. 家产 ##

### 原题链接 ###

[AcWing 1604. 家产][AcWing 1604.]

### 算法标签 ###

并查集

### 思路 ###

将关系信息读入存储， 依题意合并

##### 存储数组含义 #####

![在这里插入图片描述][5e29599b895e42e6bed8ce6ad3d57cd0.png]

##### 依题意合并操作 #####

![在这里插入图片描述][0c17de7d7fec465aa39437062dc9915f.png]

![在这里插入图片描述][8c03ee7af2d344db962c5316a1fa158b.png]

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #include<bits/stdc++.h>
    #define int long long
    #define xx first
    #define yy second
    #define ump unordered_map
    #define us unordered_set
    #define pq priority_queue
    #define rep(i, a, b) for(int i=a;i<b;++i)
    #define Rep(i, a, b) for(int i=a;i>=b;--i)
    using namespace std;
    typedef pair<int, int> PII;
    const int N=10005, inf=0x3f3f3f3f3f3f3f3f, mod=1e9+7;
    const double Exp=1e-8;
    //int t, n, m, cnt, ans; 
    int n, p[N], c[N], hc[N], ha[N], m, k, kk, id, f, mo;
    bool st[N];
    struct Ed{
        int a, b;
    }e[N];
    struct Fam{
        int id, c, hc, ha;
        bool operator<(const Fam &t) const{
            if(ha*t.c!=t.ha*c){
                return ha*t.c>t.ha*c;
            }
            return id<t.id;
        }
    };
    inline int rd(){
       int s=0,w=1;
       char ch=getchar();
       while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
       while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
    }
    void put(int x) {
        if(x<0) putchar('-'),x=-x;
        if(x>=10) put(x/10);
        putchar(x%10^48);
    }
    int find(int x){
        if(x!=p[x]){
            return p[x]=find(p[x]);
        }
        return p[x];
    }
    signed main(){
    	ios::sync_with_stdio(false);
    	cin.tie(0);
    	cout.tie(0);
    	n=rd();
    	m=0;
    	while(n--){
    	    id=rd(), f=rd(), mo=rd();
    	    st[id]=true;
    	    if(f!=-1){
    	        e[m++]={id, f};
    	    }
    	    if(mo!=-1){
    	        e[m++]={id, mo};
    	    }
    	    k=rd();
    	    while(k--){
    	        kk=rd();
    	        e[m++]={id, kk};
    	    }
    	    hc[id]=rd(), ha[id]=rd();
    	}
    	rep(i, 0, N){
    	    p[i]=i;
    	    c[i]=1;
    	}
    	rep(i, 0, m){
    	    int a=e[i].a, b=e[i].b;
    	    st[a]=st[b]=true;
    	    int pa=find(a), pb=find(b);
    	    if(pa!=pb){
    	        if(pa<pb){
    	            swap(pa, pb);
    	        }
    	        c[pb]+=c[pa];
    	        hc[pb]+=hc[pa];
    	        ha[pb]+=ha[pa];
    	        p[pa]=pb;
    	        
    	    }
    	}
    	vector<Fam> fam;
    	rep(i, 0, N){
    	    if(st[i]&&p[i]==i){
    	        fam.push_back({i, c[i], hc[i], ha[i]});
    	    }
    	}
    	sort(fam.begin(), fam.end());
    	printf("%lld\n", fam.size());
    	for(auto x:fam){
    	    printf("%04lld %lld %.3lf %.3lf\n", x.id, x.c, (double)x.hc/x.c, (double)x.ha/x.c);
    	}
    	return 0;
    }

### 参考文献 ###

[AcWing 1604. 家产（PAT甲级辅导课）y总视频讲解][AcWing 1604. _PAT_y]

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![在这里插入图片描述][9cb4cb80a18345158f8a69e2f196e5b9.png]

[AcWing 1604.]: https://www.acwing.com/problem/content/1606/
[5e29599b895e42e6bed8ce6ad3d57cd0.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/b47903a1487d43448ac7f6e1ceda398a.png
[0c17de7d7fec465aa39437062dc9915f.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/45c6cdb3d3ba414897fd5891f440f2b9.png
[8c03ee7af2d344db962c5316a1fa158b.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/ea45f4aed7c140258b3fcaad3d11a968.png
[AcWing 1604. _PAT_y]: https://www.acwing.com/video/1190/
[9cb4cb80a18345158f8a69e2f196e5b9.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/820f1ae561a24061870ff214b79d3660.png