• 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:
    分页存储过程
    WinForm中DataGridView显示更新数据--人性版
    char类型的说明
    代码创建数据库_表--SqlServer数据库
    单例设计模式
    c#中的正则表达式
    sessionStorage 和 localStorage
    图片懒加载插件lazyload.js详解
    git安装加操作(转)
    php获取数据转换成json格式
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4722380.html
Copyright © 2020-2023  润新知