codeforces 1041 E. Tree Reconstruction 和度数有关的构造树

CF 1041E：http://codeforces.com/contest/1041/problem/E

 

题意：

　　　　告诉你一个树的节点个数，显然有n-1条边。已知去掉一条边后，两个集合中最大的节点值。问原来的树形状是怎么样的，构造不出来就输出NO。

思路：

　　　　这里说的“度数”可能有点不恰当。指以这个点引出一条链的长度，链上点的值小于这个点。

　　　　我想着这应该是可以作为一条链的，但是一直没有想到向节点度数上去想。首先，输入的一对值中，有一个一定是等于n的，那另一个值我们给它度数++。我们把度数为0的点从大到小加入到队列中。然后枚举度数大于1的点，从队列中取出较为自由的点当作链上的点。注意，如果自由的最大点比当前点要大，那么肯定是不存在的。

　

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
 
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
 
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
 
 
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;
 
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'
 
#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;
 
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
// const int mod = 10007;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;
 
 
template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
 
 
/*-----------------------showtime----------------------*/
            const int maxn = 2000;
            int n;
            int du[maxn];
            queue<int>que;
            vector<pii>ans;
int main(){
            
            cin>>n;
            int flag=1,cnt = 0;
            for(int i=1; i<n; i++){
                int u,v;
                cin>>u>>v;
                if(u<n && v<n){
                    puts("NO");
                    return 0;
                }
                if(u == n)du[v]++;
                else du[u] ++;
            }
            for(int i=n-1; i>=1; i--){
                if(du[i] == 0)que.push(i);
            }
            for(int i=n-1; i>=1; i--){
                if(du[i] == 0)continue;
                du[i] --;
                int u = i;
                while(du[i]--){

                     if(que.empty()||que.front() > i){  //无法成链，NO
                        puts("NO");
                        return 0;
                     }
                     int v = que.front();que.pop();
                     ans.pb(pii(v, u));
                     u = v;
                }
                ans.pb(pii(u,n));
            }
            puts("YES");
            for(int i=0; i<ans.size(); i++){
                printf("%d %d\n", ans[i].fi, ans[i].se);
            }
            return 0;
}