• 2015 Multi-University Training Contest 7 hdu 5379 Mahjong tree


    Mahjong tree

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 388    Accepted Submission(s): 125


    Problem Description
    Little sun is an artist. Today he is playing mahjong alone. He suddenly feels that the tree in the yard doesn't look good. So he wants to decorate the tree.(The tree has n vertexs, indexed from 1 to n.)
    Thought for a long time, finally he decides to use the mahjong to decorate the tree.
    His mahjong is strange because all of the mahjong tiles had a distinct index.(Little sun has only n mahjong tiles, and the mahjong tiles indexed from 1 to n.)
    He put the mahjong tiles on the vertexs of the tree.
    As is known to all, little sun is an artist. So he want to decorate the tree as beautiful as possible.
    His decoration rules are as follows:

    (1)Place exact one mahjong tile on each vertex.
    (2)The mahjong tiles' index must be continues which are placed on the son vertexs of a vertex.
    (3)The mahjong tiles' index must be continues which are placed on the vertexs of any subtrees.

    Now he want to know that he can obtain how many different beautiful mahjong tree using these rules, because of the answer can be very large, you need output the answer modulo 1e9 + 7.
     
    Input
    The first line of the input is a single integer T, indicates the number of test cases. 
    For each test case, the first line contains an integers n. (1 <= n <= 100000)
    And the next n - 1 lines, each line contains two integers ui and vi, which describes an edge of the tree, and vertex 1 is the root of the tree.
     
    Output
    For each test case, output one line. The output format is "Case #x: ans"(without quotes), x is the case number, starting from 1.
     
    Sample Input
    2
    9
    2 1
    3 1
    4 3
    5 3
    6 2
    7 4
    8 7
    9 3
    8
    2 1
    3 1
    4 3
    5 1
    6 4
    7 5
    8 4
     
    Sample Output
    Case #1: 32
    Case #2: 16
     
    Source
     
    解题:。。。
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <vector>
     4 #include <cstring>
     5 #pragma comment(linker, "/STACK:102400000,102400000")
     6 using namespace std;
     7 const int maxn = 100010;
     8 typedef long long LL;
     9 const LL mod = 1e9 + 7;
    10 vector<int>g[maxn];
    11 int son[maxn];
    12 LL fac[maxn] = {1};
    13 LL dfs(int u,int fa){
    14     son[u] = 1;
    15     LL ret = 1;
    16     int a = 0,b = 0;
    17     for(int i = g[u].size()-1; i >= 0; --i){
    18         if(g[u][i] == fa) continue;
    19         ret = ret*dfs(g[u][i],u)%mod;
    20         if(son[g[u][i]] > 1) b++;
    21         else a++;
    22         if(!ret || b > 2) return 0;
    23         son[u] += son[g[u][i]];
    24     }
    25     if(b) ret = ret*2%mod;
    26     ret = ret*fac[a]%mod;
    27     return ret;
    28 }
    29 void init(){
    30     for(int i = 1; i < maxn; ++i)
    31         fac[i] = i*fac[i-1]%mod;
    32 }
    33 int main(){
    34     int kase,n,u,v,cs = 1;
    35     init();
    36     scanf("%d",&kase);
    37     while(kase--){
    38         scanf("%d",&n);
    39         for(int i = 0; i < maxn; ++i) {g[i].clear();son[i] = 0;}
    40         for(int i = 1; i < n; ++i){
    41             scanf("%d%d",&u,&v);
    42             g[u].push_back(v);
    43             g[v].push_back(u);
    44         }
    45         LL ret = dfs(1,-1);
    46         if(son[1] > 1) ret = ret*2%mod;
    47         printf("Case #%d: %I64d
    ",cs++,ret);
    48     }
    49     return 0;
    50 }
    View Code
  • 相关阅读:
    Docker概念学习系列之详谈Docker 的核心组件与概念(5)
    全网最详细的如何在谷歌浏览器里正确下载并安装Postman【一款功能强大的网页调试与发送网页HTTP请求的Chrome插件】(图文详解)
    全网最详细的一款满足多台电脑共用一个鼠标和键盘的工具Synergy(图文详解)
    [转]华为开发者联盟开放的服务
    [转]英语发音规则---E字母常见的发音组合有哪些
    [转]【信息系统项目管理师】重点整理:高项知识地图
    [转]【信息系统项目管理师】高项案例分析攻略
    [转]【信息系统项目管理师】案例分析记忆题
    [转]850 Basic English words
    [转]信息系统项目管理师计算题汇总
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4722380.html
Copyright © 2020-2023  润新知