Petya and Construction Set（图的构造） Codeforces Round #583 (Div. 1 + Div. 2, based on Olympiad of Metropolises)

题意：https://codeforc.es/contest/1214/problem/E

有2n个点，每个2*i和2*i-1的距离必须是Di（<=n)，现在让你构造这个树。

思路：

因为Di小于等于n，所以先对Di从大到小排序，把左端点排成一排，然后右端点搞搞就行。

注意：如果右端点应该插到最后一个点上面，那就把它变成新的最有一个点（++n）。

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 #include <cstdio>//sprintf islower isupper
 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
 #include <bitset>
 //#include <map>
 //#include<unordered_map>
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr substr
 #include <string>
 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <deque>
 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 #include <vector>//emplace_back
 //#include <math.h>
 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 #define fo(a,b,c) for(register int a=b;a<=c;++a)
 #define fr(a,b,c) for(register int a=b;a>=c;--a)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define pr printf
 #define sc scanf
 #define ls rt<<1
 #define rs rt<<1|1
 typedef long long ll;
 void swapp(int &a,int &b);
 double fabss(double a);
 int maxx(int a,int b);
 int minn(int a,int b);
 int Del_bit_1(int n);
 int lowbit(int n);
 int abss(int a);
 //const long long INF=(1LL<<60);
 const double E=2.718281828;
 const double PI=acos(-1.0);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e6+;

 struct node
 {
     int id,len;
     friend bool operator<(node a,node b)
     {
         return a.len>b.len;
     }
 }edge[N];
 vector<vector<node> >v(N);

 int main()
 {
     int n;
     sc("%d",&n);
     for(int i=;i<=n;++i)
         sc("%d",&edge[i].len),edge[i].id=i*-;
     sort(edge+,edge++n);
     for(int i=;i<=n;++i)
         v[i].push_back(edge[i]);
     int End=n;
     for(int i=;i<=n;++i)
     {
         node temp=v[i][];
         if(temp.len+i<=End)
             v[i+temp.len-].push_back({temp.id+,});
         else
             v[++End].push_back({temp.id+,});
     }
     for(int i=;i<=End;++i)
     {
         if(i!=)
             pr("%d %d\n",v[i-][].id,v[i][].id);
         int sz=v[i].size();
         for(int j=;j<sz;++j)
             pr("%d %d\n",v[i][].id,v[i][j].id);
     }
     return ;
 }

 /**************************************************************************************/

 int maxx(int a,int b)
 {
     return a>b?a:b;
 }

 void swapp(int &a,int &b)
 {
     a^=b^=a^=b;
 }

 int lowbit(int n)
 {
     return n&(-n);
 }

 int Del_bit_1(int n)
 {
     return n&(n-);
 }

 int abss(int a)
 {
     return a>?a:-a;
 }

 double fabss(double a)
 {
     return a>?a:-a;
 }

 int minn(int a,int b)
 {
     return a<b?a:b;
 }