最短路。

每次询问的点当做起点,然后算一下点到其余点的最短路。然后统计一下最短路小于等于L的点有几个。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std; const int INF=0x7FFFFFFF;
const int maxn=+;
struct Edge
{
int u,v;
int dis;
}e[*maxn];
vector<int>g[maxn];
int tot;
int flag[maxn],dis[maxn];
int n,L; void SPFA(int s)
{
queue<int>Q;
memset(flag,,sizeof flag);
for(int i=;i<=n;i++) dis[i]=INF;
Q.push(s); flag[s]=; dis[s]=;
while(!Q.empty())
{
int head=Q.front(); Q.pop(); flag[head]=;
for(int i=;i<g[head].size();i++)
{
int id=g[head][i];
if(dis[head]+e[id].dis<dis[e[id].v])
{
dis[e[id].v]=dis[head]+e[id].dis;
if(flag[e[id].v]==)
{
flag[e[id].v]=;
Q.push(e[id].v);
}
}
}
}
} int main()
{
tot=;
scanf("%d%d",&n,&L);
for(int i=;i<=n;i++)
{
int m; scanf("%d",&m);
while(m--)
{
int id; scanf("%d",&id);
e[tot].u=id; e[tot].v=i; e[tot].dis=;
g[id].push_back(tot);
tot++;
}
}
int k; scanf("%d",&k);
for(int i=;i<=k;i++)
{
int id; scanf("%d",&id);
SPFA(id);
int ans=;
for(int i=;i<=n;i++)
if(i!=id&&dis[i]<=L) ans++;
printf("%d\n",ans);
}
return ;
}

PAT (Advanced Level) 1076. Forwards on Weibo (30)的更多相关文章

  1. 【PAT甲级】1076 Forwards on Weibo (30 分)

    题意: 输入两个正整数N和L(N<=1000,L<=6),接着输入N行数据每行包括它关注人数(<=100)和关注的人的序号,接着输入一行包含一个正整数K和K个序号.输出每次询问的人发 ...

  2. 1076. Forwards on Weibo (30)【树+搜索】——PAT (Advanced Level) Practise

    题目信息 1076. Forwards on Weibo (30) 时间限制3000 ms 内存限制65536 kB 代码长度限制16000 B Weibo is known as the Chine ...

  3. PAT 甲级 1076 Forwards on Weibo (30分)(bfs较简单)

    1076 Forwards on Weibo (30分)   Weibo is known as the Chinese version of Twitter. One user on Weibo m ...

  4. PAT Advanced 1076 Forwards on Weibo (30) [图的遍历,BFS,DFS]

    题目 Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and ...

  5. PAT 1076. Forwards on Weibo (30)

    Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may ...

  6. PAT甲题题解-1076. Forwards on Weibo (30)-BFS

    题目大意:给出每个用户id关注的人,和转发最多的层数L,求一个id发了条微博最多会有多少个人转发,每个人只考虑转发一次.用BFS,同时每个节点要记录下所在的层数,由于只能转发一次,所以每个节点要用vi ...

  7. 1076. Forwards on Weibo (30)

    时间限制 3000 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Weibo is known as the Chinese v ...

  8. 1076. Forwards on Weibo (30) - 记录层的BFS改进

    题目如下: Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, a ...

  9. 1076 Forwards on Weibo (30)(30 分)

    Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may ...

随机推荐

  1. Educational Codeforces Round 15_D. Road to Post Office

    D. Road to Post Office time limit per test 1 second memory limit per test 256 megabytes input standa ...

  2. 期刊:DOI :10.3969/j.issn.1001-2400.2012.01.008

    DOI:10.3969/j.issn.1001-2400.2012.01.008 “/” 分为前缀和后缀两部分中间用一斜杠分开,前缀中又以小圆点分为两部分. "DOI":一篇期刊论 ...

  3. 转载--C# PLINQ 内存列表查询优化历程

    http://www.cnblogs.com/dengxi/p/5305066.html 产品中(基于ASP.NET MVC开发)需要经常对药品名称及名称拼音码进行下拉匹配及结果查询.为了加快查询的速 ...

  4. 上传下载文件, 同时部署在webapps下, 而不是项目下,防止重新部署tomcat, 上传文件消失

    前端上传 <a href='javascript:upload("+data[i].id+")' title='Upload Report'> <img src= ...

  5. ubuntu 自动获取ip的怎么设置

    ubuntu以DHCP方式配置网卡自动获取ip编辑文件/etc/network/interfaces:sudo vi /etc/network/interfaces并用下面的行来替换有关eth0的行: ...

  6. sdio/sd/mmc architecture

    三个驱动mmc_core.ko mmc_block.ko omap_hsmmc.ko 读懂一个协议的架构有利于学习源码 参考了 http://processors.wiki.ti.com/index. ...

  7. Mybatis 一对一,一对多,多对一,多对多的理解

    First (一对一) 首先我来说下一对一的理解,就是一个班主任只属于一个班级,一个班级也只能有一个班主任.好吧这就是对于一对一的理解 怎么来实现呢? 这里我介绍了两种方式: 一种是:使用嵌套结果映射 ...

  8. TexturePacker license Key免费获取方式

    TexturePacker是一款功能非常强大的图片制作工具.是一款付费软件,但是TexturePacker的作者Andreas Löw先生也给出获得免费key 的方法...大家可以到这个网站去申请 h ...

  9. Android自定义XML属性

    <?xml version="1.0" encoding="utf-8"?> <resources> <declare-style ...

  10. 深入解析FileInputStream和FileOutputStream

    http://swiftlet.net/archives/1363 FileInputStream和FileOutputStream类属于字节类,可以操作任意类型的文件.在数据流的处理过程中,有两种情 ...