Destroying the bus stations

                                                                                    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

                                                                                                           Total Submission(s): 2072    Accepted Submission(s): 647

Problem Description
                Gabiluso is one of the greatest spies in his country. Now he’s trying to complete an “impossible” mission ----- to make it slow for the army of City Colugu to reach the airport. City Colugu has n bus stations and m roads. Each road connects two bus stations directly, and all roads are one way streets. In order to keep the air clean, the government bans all military vehicles. So the army must take buses to go to the airport. There may be more than one road between two bus stations. If a bus station is destroyed, all roads connecting that station will become no use. What’s Gabiluso needs to do is destroying some bus stations to make the army can’t get to the airport in k minutes. It takes exactly one minute for a bus to pass any road. All bus stations are numbered from 1 to n. The No.1 bus station is in the barrack and the No. n station is in the airport. The army always set out from the No. 1 station. No.1 station and No. n station can’t be destroyed because of the heavy guard. Of course there is no road from No.1 station to No. n station.
Please help Gabiluso to calculate the minimum number of bus stations he must destroy to complete his mission.
 
Input
     There are several test cases. Input ends with three zeros.
      For each test case:
     The first line contains 3 integers, n, m and k. (0< n <=50, 0< m<=4000, 0 < k < 1000) Then m lines follows.
     Each line contains 2 integers, s and f, indicating that there is a road from station No. s to station No. f. 
 
Output
For each test case, output the minimum number of stations Gabiluso must destroy.
 
Sample Input
5 7 3
1 3
3 4
4 5
1 2
2 5
1 4
4 5
0 0 0
 
Sample Output
2
 
Source
 
此题是最大最小流/现在还在温习中....!
题意:
Gabiluso是
 贴一下
代码:

 #include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm> using namespace std; const int maxm = ; //最大边数
const int maxn = ; //最大点数 struct aaa
{
int s,f,next ;
}; aaa c[maxm];
int sta[maxn],fa[maxn],zh[maxn];
int d[maxn][maxn],e[maxn];
bool b[maxn];
int n,m,now,tot;
bool goal; void ins(int s ,int f) //创建邻接表
{
now++;
c[now].s=s;
c[now].f=f;
c[now].next=sta[s];
sta[s]=now;
} void bfs()
{
int i,cl,op,k,t;
cl=;op=;
for(i=;i<=n ;i++) fa[i]=;
zh[]=;fa[]=-;
while(cl<op)
{
cl++;
k=zh[cl];
for( t=sta[k] ; t ; t=c[t].next )
if(b[c[t].f]&&fa[c[t].f]==)
{
op++;
zh[op]=c[t].f;
fa[c[t].f]=c[t].s;
if(c[t].f==n) break;
}
if(fa[n]) break ;
}
} void dfs(int deep)
{
int i,cl,op,l,k;
if(goal) return ;
bfs();
if(fa[n]==)
{
goal=true ;
return ;
}
l=;
for(k=n ;k>l ; k=fa[k])
{
l++;
d[deep][l]=k;
}
if(l>m)
{
goal=true;
return ;
}
if(deep>tot) return ; for(i=;i<=l;i++)
{
b[d[deep][i]]=false ;
if(e[d[deep][i]]==) dfs(deep+);
b[d[deep][i]]=true;
e[d[deep][i]]++;
}
for(i= ; i<=l ; i++ )
e[d[deep][i]]--;
} int make()
{
int i,j;
goal= false;
for(i= ;i<=n ;i++)
{
tot=i;
for(j=;j<=n ;j++) b[j]=true;
memset(e,,sizeof(e));
dfs();
if(goal) return i;
}
return n;
} int main()
{
int i,s,f,g;
while()
{
scanf("%d%d%d",&n,&g,&m);
if(n==) break;
memset(sta,,sizeof(sta));
now=;
for(i=; i<=g ;i++)
{
scanf("%d%d",&s,&f);
ins(s,f);
}
g=make();
printf("%d\n",g);
}
return ;
}
 

HDUOJ----2485 Destroying the bus stations(2008北京现场赛A题)的更多相关文章

  1. HDUOJ-------2493Timer(数学 2008北京现场赛H题)

    Timer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Subm ...

  2. HDUOJ--------A simple stone game(尼姆博弈扩展)(2008北京现场赛A题)

    A simple stone game                                                                                  ...

  3. HDU 2485 Destroying the bus stations(!最大流∩!费用流∩搜索)

    Description Gabiluso is one of the greatest spies in his country. Now he’s trying to complete an “im ...

  4. 图论--网络流--最小割 HDU 2485 Destroying the bus stations(最短路+限流建图)

    Problem Description Gabiluso is one of the greatest spies in his country. Now he's trying to complet ...

  5. HDU 2485 Destroying the bus stations (IDA*+ BFS)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2485 题意:给你n个点,m条相连的边,问你最少去掉几个点使从1到n最小路径>=k,其中不能去掉1, ...

  6. HDU 2485 Destroying the bus stations(费用流)

    http://acm.hdu.edu.cn/showproblem.php?pid=2485 题意: 现在要从起点1到终点n,途中有多个车站,每经过一个车站为1时间,现在要在k时间内到达终点,问至少要 ...

  7. HDU 2485 Destroying the bus stations

    2015 ACM / ICPC 北京站 热身赛 C题 #include<cstdio> #include<cstring> #include<cmath> #inc ...

  8. hdu 2485 Destroying the bus stations 最小费用最大流

    题意: 最少需要几个点才能使得有向图中1->n的距离大于k. 分析: 删除某一点的以后,与它相连的所有边都不存在了,相当于点的容量为1.但是在网络流中我们只能直接限制边的容量.所以需要拆点来完成 ...

  9. hdu-----2491Priest John's Busiest Day(2008 北京现场赛G)

    Priest John's Busiest Day Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Jav ...

随机推荐

  1. linux 中 ll 命令如何让查询结果按时间升序或降序排序?

    -t选项的功能是使输出的结果将以时间降序排列.如果希望按时间的升序排列,可以使用管道符将返回的结果传入tac命令.用法示例:查询当前目录的文件并以降序排列: ll -t查询当前目录的文件并以升序排列: ...

  2. js跨域及解决方案

    本文出自:http://www.cnblogs.com/oneword/archive/2012/12/03/2799443.html 1.什么是跨域 我们经常会在页面上使用ajax请求访问其他服务器 ...

  3. 哈理工软件学院"兆方美迪"杯第六届程序设计大赛【高年级组】--决赛 题解

    比赛链接:http://acm-software.hrbust.edu.cn/contest.php?cid=1082 A.好SB啊真是,还以为lis-数有多少个数不一样. #include < ...

  4. linux挂载共享文件夹

    挂载windows共享目录或FTP: 方式一:包含密码 Shell代码 收藏代码 sudo mount //192.168.10.22/FTPServer /windows -o username=u ...

  5. react入门笔记

    this.props.children是任何内嵌的元素 利用ref属性给子组件命名,this.refs引用组件,getDOMNode()获取本地的DOM元素,如: this.refs.author.g ...

  6. Python基础学习笔记(八)常用字典内置函数和方法

    参考资料: 1. <Python基础教程> 2. http://www.runoob.com/python/python-dictionary.html 3. http://www.lia ...

  7. 2013/7/17 HNU_训练赛5

    sgu 542 Gena vs Petya sgu 543 Cafe 题意:有N组人需要被分配到某些固定了人数的桌子上,其中ai表示第i组有多少个人,安排作为需要符合如下安排:某一组的人员不能够单独在 ...

  8. iOS - UIScrollView

    前言 NS_CLASS_AVAILABLE_IOS(2_0) @interface UIScrollView : UIView <NSCoding> @available(iOS 2.0, ...

  9. JAVA 大数据内存耗用测试

    JAVA 大数据内存耗用测试import java.lang.management.ManagementFactory;import java.lang.management.MemoryMXBean ...

  10. [转载] Linux下高并发socket最大连接数所受的各种限制

    原文: http://mp.weixin.qq.com/s?__biz=MzAwNjMxNjQzNA==&mid=207772333&idx=1&sn=cfc8aadb422f ...