微软2016校园招聘在线笔试 B Professor Q's Software [ 拓扑图dp ]
题目2 : Professor Q's Software
描述
Professor Q develops a new software. The software consists of N modules which are numbered from 1 to N. The i-th module will be started up by signal Si. If signal Si is generated multiple times, the i-th module will also be started multiple times. Two different modules may be started up by the same signal. During its lifecircle, the i-th module will generate Ki signals: E1, E2, ..., EKi. These signals may start up other modules and so on. Fortunately the software is so carefully designed that there is no loop in the starting chain of modules, which means eventually all the modules will be stoped. Professor Q generates some initial signals and want to know how many times each module is started.
输入
The first line contains an integer T, the number of test cases. T test cases follows.
For each test case, the first line contains contains two numbers N and M, indicating the number of modules and number of signals that Professor Q generates initially.
The second line contains M integers, indicating the signals that Professor Q generates initially.
Line 3~N + 2, each line describes an module, following the format S, K, E1, E2, ... , EK. S represents the signal that start up this module. K represents the total amount of signals that are generated during the lifecircle of this module. And E1 ... EK are these signals.
For 20% data, all N, M <= 10
For 40% data, all N, M <= 103
For 100% data, all 1 <= T <= 5, N, M <= 105, 0 <= K <= 3, 0 <= S, E <= 105.
Hint: HUGE input in this problem. Fast IO such as scanf and BufferedReader are recommended.
输出
For each test case, output a line with N numbers Ans1, Ans2, ... , AnsN. Ansi is the number of times that the i-th module is started. In case the answers may be too large, output the answers modulo 142857 (the remainder of division by 142857).
- 样例输入
-
3
3 2
123 256
123 2 456 256
456 3 666 111 256
256 1 90
3 1
100
100 2 200 200
200 1 300
200 0
5 1
1
1 2 2 3
2 2 3 4
3 2 4 5
4 2 5 6
5 2 6 7 - 样例输出
-
1 1 3
1 2 2
1 1 2 3 5
题意:
一个有向无环图,初始访问某些点,访问过的点会沿着能连的边一直走到底,问,最后每个点分别被访问了几次。
题解:
来自天猫的思路。
拓扑图dp。一个很好的思路~~~
从根节点开始,如果某个节点访问了,它的所有儿子节点访问数+1。
由于是按照拓扑顺序来处理的(并且没有环),所以,在继续对儿子的儿子处理时,不会再出现儿子节点再增加访问数。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <set>
#include <utility>
#include <vector>
#include <queue> using namespace std; typedef pair<int,int> PII;
typedef pair<int,PII> PIII; #define LL long long
#define ULL unsigned long long
#define m_p make_pair
#define l_b lower_bound
#define p_b push_back
#define w1 first
#define w2 second
#define maxlongint 2147483647
#define biglongint 2139062143 const int maxn=;
const int A=; int TT,N,M,o,sc,tj;
vector<int> F[maxn];
int c[maxn],a[maxn],ans[maxn],vis[maxn],dp[maxn],inp[maxn]; void dfs(int s)
{
if (vis[s]==) return;
vis[s]=;
for (int i=;i<F[s].size();i++)
dfs(F[s][i]);
++o,ans[o]=s;
} int main()
{
scanf("%d",&TT);
for (int gb=;gb<=TT;gb++)
{
scanf("%d %d",&N,&M);
for (int i=;i<=M;i++) scanf("%d",&c[i]);
memset(inp,,sizeof(inp));
for (int i=;i<=A;i++) F[i].clear();
for (int i=;i<=N;i++)
{
scanf("%d",&a[i]);
scanf("%d",&sc);
for (int j=;j<=sc;j++)
{
scanf("%d",&tj);
if (tj>A) continue;
F[a[i]].p_b(tj);
++inp[tj];
}
}
o=;
memset(vis,,sizeof(vis));
for (int i=;i<=A;i++)
if (inp[i]==) dfs(i);
memset(dp,,sizeof(dp));
for (int i=;i<=M;i++) dp[c[i]]++;
for (int i=A+;i>=;i--)
{
sc=ans[i];
for (int j=;j<F[sc].size();j++)
dp[F[sc][j]]+=dp[sc],dp[F[sc][j]]%=;
}
for (int i=;i<N;i++) printf("%d ",dp[a[i]]);printf("%d\n",dp[a[N]]);
}
return ;
}
微软2016校园招聘在线笔试 B Professor Q's Software [ 拓扑图dp ]的更多相关文章
- 微软2016校园招聘在线笔试-Professor Q's Software
题目2 : Professor Q's Software 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 Professor Q develops a new softw ...
- 微软2016校园招聘在线笔试第二场 题目1 : Lucky Substrings
时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 A string s is LUCKY if and only if the number of different ch ...
- 微软2016校园招聘在线笔试 [Recruitment]
时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 A company plans to recruit some new employees. There are N ca ...
- 题目3 : Spring Outing 微软2016校园招聘在线笔试第二场
题目3 : Spring Outing 时间限制:20000ms 单点时限:1000ms 内存限制:256MB 描述 You class are planning for a spring outin ...
- 微软2016校园招聘在线笔试之Magic Box
题目1 : Magic Box 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 The circus clown Sunny has a magic box. When ...
- hihocoder 1288 : Font Size (微软2016校园招聘4月在线笔试)
hihocoder 1288 笔试第一道..wa了好几次,也是无语..hihocoder错了不会告诉你失败的时候的测试集,这样有时候就很烦.. 遍历所有的字体,从min(w,h)开始逐渐变小开始遍历. ...
- 微软2016校园招聘4月在线笔试 A FontSize
题目链接:http://hihocoder.com/problemset/problem/1288 分析:题目中所求的是最大的FontSize(记为S),其应该满足P*[W/S]*[H/S] > ...
- 微软2016校园招聘4月在线笔试 ABC
题目链接:http://hihocoder.com/contest/mstest2016april1/problems 第一题:输入N,P,W,H,代表有N段文字,每段有ai个字,每行有⌊W/S⌋个字 ...
- 微软2016校园招聘4月在线笔试 hihocoder 1289 403 Forbidden
时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描写叙述 Little Hi runs a web server. Sometimes he has to deny acces ...
随机推荐
- 新手玩CSS中的一些黑科技
哎哎 1.鼠标移进网页里,不见了= = *{ cursor: none!important; } 2.简单的文字模糊效果 *{ color: transparent; text-shadow: #11 ...
- Android(java)学习笔记181:多媒体之图片画画板案例
1.首先我们编写布局文件activity_main.xml如下: <RelativeLayout xmlns:android="http://schemas.android.com/a ...
- Swift学习——流程控制
1.for in循环 (1)简单使用: for-in和范围运算符 for i in 1...3 { println(i) } (2)如果在循环中用不到i,可用_代替 for _ in 1...3 { ...
- Java中的代理--proxy
讲到代理,好像在之前的springMVC,还是spring中或者是hibernate中学习过,并没有特别在意,这次好好理解一下.(原来是在spring中的AOP,面向切面 Aspect Oriente ...
- excel数据比对,查找差异
1.选中需比对的数据 2.开始->条件格式->突出显示单元格规则->重复值 3.选择唯一值,点击确定 4.结果展示 5.颜色标识的即:不同值
- Java编程:常见问题汇总
每天在写Java程序,其实里面有一些细节大家可能没怎么注意,这不,有人总结了一个我们编程中常见的问题.虽然一般没有什么大问题,但是最好别这样做. AD: 每天在写Java程序,其实里面有一些细节大家可 ...
- vue在传值的时候经常遇到的问题
在我用vue编写程序的时候,在传值的时候,经常会遇到些问题,像今天遇到了两个问题,在用父传子的方法去传值,当父组件中的要传的数据是for循环出来的或者是列表的时候,你想每次运行的事件,都去传某一行,或 ...
- bzoj 3555 企鹅QQ
https://www.lydsy.com/JudgeOnline/problem.php?id=3555 枚举每一位字符,计算字符两侧的哈希值,然后进行比较,用map或排序记录出与其相同的字符串数量 ...
- Day02:我的Python学习之路
1.初识模块 Python的强大之处在于他有非常丰富和强大的标准库和第三方库,现在简单的学习2个常见的标准库——sys和os. (1)系统的标准库sys # Author:GCL # 系统的标准库sy ...
- 2018美赛准备之路——Matlab基础——基本运算符号表示
π pi ln(x) log(x) lg(x) log10(x) log2(x) log2(x) 根号 sqrt(x) x的y次方 x^y e的y次方 exp(y)