zoukankan      html  css  js  c++  java
  • Codeforces #541 (Div2)

    Problem   Codeforces #541 (Div2) - F. Asya And Kittens

    Time Limit: 2000 mSec

    Problem Description

    Input

    The first line contains a single integer nn (2n150000) — the number of kittens.

    Each of the following n1lines contains integers xi and yi (1xi,yin, xiyi) — indices of kittens, which got together due to the border removal on the corresponding day.

    It's guaranteed, that the kittens xi and yi were in the different cells before this day.

    Output

    For every cell from 1 to n print a single integer — the index of the kitten from 1 to n, who was originally in it.

    All printed integers must be distinct.

    It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.

    Sample Input

    5
    1 4
    2 5
    3 1
    4 5

    Sample Output

    3 1 4 2 5

    题解:并查集+链表,始终把y所在的集合加到x所在集合的右边,让每个集合的根始终保持在最左边,维护一下每个集合最左边元素,链表连一连即可,这种题拼的就是手速。

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 #define REP(i, n) for (int i = 1; i <= (n); i++)
     6 #define sqr(x) ((x) * (x))
     7 
     8 const int maxn = 150000 + 100;
     9 const int maxm = 200000 + 100;
    10 const int maxs = 10000 + 10;
    11 
    12 typedef long long LL;
    13 typedef pair<int, int> pii;
    14 typedef pair<double, double> pdd;
    15 
    16 const LL unit = 1LL;
    17 const int INF = 0x3f3f3f3f;
    18 const LL mod = 1000000007;
    19 const double eps = 1e-14;
    20 const double inf = 1e15;
    21 const double pi = acos(-1.0);
    22 
    23 int n;
    24 int fa[maxn], ri[maxn];
    25 int ans[maxn];
    26 int edge[maxn];
    27 
    28 int findn(int x)
    29 {
    30     return x == fa[x] ? x : fa[x] = findn(fa[x]);
    31 }
    32 
    33 void init()
    34 {
    35     for (int i = 0; i < maxn; i++)
    36     {
    37         fa[i] = edge[i] = ri[i] = i;
    38     }
    39 }
    40 
    41 int main()
    42 {
    43     ios::sync_with_stdio(false);
    44     cin.tie(0);
    45     //freopen("input.txt", "r", stdin);
    46     //freopen("output.txt", "w", stdout);
    47     init();
    48     cin >> n;
    49     int x, y;
    50     for (int i = 0; i < n - 1; i++)
    51     {
    52         cin >> x >> y;
    53         int fx = findn(x), fy = findn(y);
    54         fa[fy] = fx;
    55         ri[edge[fx]] = fy;
    56         edge[fx] = edge[fy];
    57     }
    58     int root = -1;
    59     for (int i = 1; i <= n; i++)
    60     {
    61         if (findn(i) == i)
    62         {
    63             root = i;
    64         }
    65     }
    66     ans[1] = root;
    67     int cnt = 1;
    68     for (int i = ri[root]; cnt++; i = ri[i])
    69     {
    70         ans[cnt] = i;
    71         if (cnt == n)
    72             break;
    73     }
    74     for(int i = 1; i <= n; i++)
    75     {
    76         cout << ans[i];
    77         if(i != n)
    78         {
    79             cout << " ";
    80         }
    81         else
    82         {
    83             cout << endl;
    84         }
    85         
    86     }
    87     return 0;
    88 }
  • 相关阅读:
    基于VB6.0的MICAPS风云二号卫星云图转化实例(转载)
    .CS文件编译生成.DLL文件 .EXE文件(C#网络搜集)(转)
    SQL SERVER 2005及以上查看各表的记录数及占空间大小
    sql2008生成insert语句
    jdk chm文档下载地址
    source insight 解决自动缩进 和 TAB键=4个SPACE
    Hibernate的Criteria的使用
    java多线程协作: wait/notifyAll ( Cooperation between tasks )
    Eclipse 去掉JavaScript Validator
    jquery 插件示例, jquery popup 插件
  • 原文地址:https://www.cnblogs.com/npugen/p/10780000.html
Copyright © 2011-2022 走看看