poj 1144 Network(割点入门)

Network

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 10907 Accepted: 5042

Description

A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.

Input

The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;

Output

The output contains for each block except the last in the input file one line containing the number of critical places.

Sample Input
5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0
Sample Output
1
2
Hint

You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.

Source

Central Europe 1996

模板～

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <algorithm>
  6 #define ll long long
  7 using namespace std;
  8 const int MAXN = 12000;
  9 const int MAXM = 120000;
 10 struct Edge
 11 {
 12     int to,next;
 13     bool cut;
 14 } edge[MAXM];
 15 int head[MAXN],tot;
 16 int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
 17 int Index,top;
 18 bool Instack[MAXN],cut[MAXN];
 19 int bridge,add_block[MAXN];
 20 
 21 void addedge(int u,int v)
 22 {
 23     edge[tot].to = v;
 24     edge[tot].next = head[u];
 25     edge[tot].cut = false;
 26     head[u] = tot++;
 27 }
 28 void Tarjan(int u,int pre)
 29 {
 30     int v;
 31     Low[u] = DFN[u] = ++Index;
 32     Stack[top++] = u;
 33     Instack[u] = true;
 34     int son = 0;
 35     for(int i = head[u]; i != -1; i = edge[i].next)
 36     {
 37         v = edge[i].to;
 38         if(v == pre)
 39             continue;
 40         if( !DFN[v] )
 41         {
 42             son++;
 43             Tarjan(v,u);
 44             if(Low[u] > Low[v]) Low[u] = Low[v];
 45 
 46             if(Low[v] > DFN[u])
 47             {
 48                 bridge++;
 49                 edge[i].cut = true;
 50                 edge[i^1].cut = true;
 51             }
 52 
 53             if(u != pre && Low[v] >= DFN[u])
 54             {
 55                 cut[u] = true;
 56                 add_block[u]++;
 57             }
 58         }
 59         else if(Instack[v] && Low[u] >  DFN[v])
 60             Low[u] = DFN[v];
 61     }
 62     if(u == pre && son > 1) cut[u] = true;
 63     if(u == pre) add_block[u] = son -1;
 64     Instack[u] = false;
 65     top--;
 66 }
 67 void solve(int N)
 68 {
 69     memset(DFN,0,sizeof(DFN));
 70     memset(Instack,false,sizeof(Instack));
 71     memset(add_block,0,sizeof(add_block));
 72     memset(cut,false,sizeof(cut));
 73     Index = top = bridge =  0;
 74     for(int i = 1; i <= N; i++)
 75         if( !DFN[i])
 76             Tarjan(i,i);
 77     int ans = 0;
 78     for(int i = 1; i <= N; i++)
 79         if(cut[i])
 80             ans++;
 81     printf("%d
",ans);
 82 }
 83 void init()
 84 {
 85     tot = 0;
 86     memset(head,-1,sizeof(head));
 87 }
 88 int main(void)
 89 {
 90     int n;
 91     int maze[120][120];
 92     while(scanf("%d",&n),n)
 93     {
 94         char ss[3005];
 95         init();
 96         memset(maze,0,sizeof(maze));
 97         getchar();
 98         while(cin.getline(ss,3000),ss[0] != '0')
 99         {
100             int u = 0,cur = 0,l = (int)strlen(ss);
101             while(ss[cur] != ' ' && cur < l)
102             {
103                 u*= 10;
104                 u += ss[cur] - '0';
105                 cur++;
106             }
107             int v;
108             while(ss[cur] && cur < l)
109             {
110                 v = 0;
111                 while(ss[cur] != ' ' && cur < l)
112                 {
113                     v*= 10;
114                     v += ss[cur] - '0';
115                     cur++;
116                 }
117                 maze[u][v] = maze[v][u] = 1;
118                 cur++;
119             }
120         }
121         for(int i = 1; i <= n; i++)
122             for(int j = 1; j < i; j++)
123                 if(maze[i][j])
124                 {
125                     addedge(i,j);
126                     addedge(j,i);
127                 }
128         solve(n);
129     }
130     return 0;
131 }

相关阅读:
【MPI学习2】MPI并行程序设计模式：对等模式 & 主从模式
 【MPI学习1】简单MPI程序示例
 【多线程】零碎记录1
【APUE】Chapter17 Advanced IPC & sign extension & 结构体内存对齐
 php-7.1编译记录
 rsyslogd系统日志服务总结
 php配置(php7.3)
php-fpm回顾和总结
 php-fpm配置项
 Hive之执行计划分析(explain)
原文地址：https://www.cnblogs.com/henserlinda/p/4782059.html

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10907		Accepted: 5042

poj 1144 Network(割点 入门)

poj 1144 Network(割点入门)