并查集 | 1114 先用并查集划分再处理数据

这题一拿到手当做BFS来做了。做到后面不能得到样例结果，冷静一分析，发现应该用并查集来做。经过这道题。我对并查集的理解更加深入了。

AC代码：

#include <stdio.h>
#include <memory.h>
#include <math.h>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <map>


#define I scanf
#define OL puts
#define O printf
#define F(a,b,c) for(a=b;a<c;a++)
#define FF(a,b) for(a=0;a<b;a++)
#define FG(a,b) for(a=b-1;a>=0;a--)
#define LEN 10010
#define MAX (1<<30)+1
#define V vector<int>

using namespace std;

int fa[LEN] ;
int findFa(int x){
    if(fa[x]==x) return x;
    int r=x;
    while(fa[r]!=r){
        r=fa[r];
    }
    int t=x;
    while(fa[x]!=x){
        t=fa[x];
        fa[x]=r;
        x=t;
    }
    return r;
}
void Union(int a,int b){//让集合b并到集合a上 
    int f_a=findFa(a);
    int f_b=findFa(b);
    fa[f_b]=f_a;
}
void init(){
    int i;
    FF(i,LEN) fa[i]=i;
}

int N,M;
struct Node{
    int id,fa=-1,mo=-1;
    vector<int> c;
    int n=0,a=0;    //num, area 
}nd[LEN]; 
vector<int> inputId;
int vis[LEN];
struct oNode{
    int id,m;
    double n,a;
};
vector<oNode> output;
bool cmp(oNode a,oNode b){
    if(a.a!=b.a) return a.a>b.a;
    return a.id<b.id;
}
set<int> idSet;
int cnt[LEN];
int area[LEN];
int estate[LEN];
int minID[LEN];

int main(){
//    freopen("I:\pat\并查集\1114.txt","r",stdin);
    fill(minID,minID+LEN,MAX);
    I("%d",&N);
    int id,t,i,j;
    init();
    while(N--){
        I("%d",&id);
        nd[id].id=id;
        inputId.push_back(id);
        I("%d%d",&nd[id].fa,&nd[id].mo);
        I("%d",&M);
        while(M--){
            I("%d",&t);
            nd[id].c.push_back(t);
        }
        I("%d%d",&nd[id].n,&nd[id].a);
    }
    FF(i,inputId.size()){
        int id=inputId[i];
        int father=nd[id].fa;
        if(father>=0) Union(id,father);
        int mother=nd[id].mo;
        if(mother>=0) Union(id,mother);
        vector<int>& cv=nd[id].c;
        FF(j,cv.size()){
            int cd=cv[j];
            if(cd>=0) Union(id,cd); 
        }
    }
    FF(i,inputId.size()){
        int id=inputId[i];
        int f=findFa(id);
        idSet.insert(f);
        //统计人数
        if(vis[id]==0){
            vis[id]=1;
            cnt[f]++;
            minID[f]=min(minID[f],id);
        }
        int father=nd[id].fa;
        if(father>=0 && vis[father]==0){
            cnt[f]++;
            vis[father]=1;
            minID[f]=min(minID[f],father);
        }
        int mother=nd[id].mo;
        if(mother>=0 && vis[mother]==0){
            cnt[f]++;
            vis[mother]=1;
            minID[f]=min(minID[f],mother);
        }
        vector<int>& cv=nd[id].c;
        FF(j,cv.size()){
            int cd=cv[j];
            if(cd>=0 && vis[cd]==0){
                Union(f,cd);
                cnt[f]++;
                vis[cd]=1;
                minID[f]=min(minID[f],cd);
            }
        }
        //增加财产
        area[f]+=nd[id].a;
        estate[f]+=nd[id].n;
    }
    set<int>::iterator it=idSet.begin();
    while(it!=idSet.end()){
        int id=*it;
        oNode tmp;
        tmp.id=minID[id];
        tmp.m=cnt[id];
        tmp.a=(double)area[id]/(double)tmp.m;
        tmp.n=(double)estate[id]/(double)tmp.m;
        output.push_back(tmp);
        it++;
    }
    sort(output.begin(),output.end(),cmp);
    O("%d
",output.size());
    FF(i,output.size()){
        O("%04d %d %.3f %.3f
",output[i].id,output[i].m,output[i].n,output[i].a);
    }
    return 0;
}