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 << " "; #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 ' ' #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 ", ans[i].fi, ans[i].se); } return 0; }