zoukankan      html  css  js  c++  java
  • 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 << "
    ";
    #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;
    }
    CF 1041E
  • 相关阅读:
    c#2005中的各个控件转换为html代码
    支付宝接口参数详谈
    IE6兼容菜单
    cookie版购物车
    火狐执行子页面方法
    easyui中datebox文本框输入非数字报错的改善
    右侧悬浮菜单
    内边距、边框和外边距
    自我超越
    DateUtils 时间工具类
  • 原文地址:https://www.cnblogs.com/ckxkexing/p/9736126.html
Copyright © 2011-2022 走看看