POJ - 1523 SPF

SPF

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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Description

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible. 

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate. 

Input

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

Output

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist. 

The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.

Sample Input

1 2
5 4
3 1
3 2
3 4
3 5
0

1 2
2 3
3 4
4 5
5 1
0

1 2
2 3
3 4
4 6
6 3
2 5
5 1
0

0

Sample Output

Network #1
  SPF node 3 leaves 2 subnets

Network #2
  No SPF nodes

Network #3
  SPF node 2 leaves 2 subnets
  SPF node 3 leaves 2 subnets

Source

Greater New York 2000

比较简单的 tarjan算法题。（待解决，模板）

/*
Memory 1172K
Time    16MS 
*/
#include <iostream>
using namespace std;
#define MAXV 1010
#define min(a,b) (a>b?b:a)


bool map[MAXV][MAXV]; //保存图
int rcnt; //记录输入的结点
int dfn[MAXV],low[MAXV]; //深度与最低深度优先数
bool vis[MAXV]; //标记结点是否被寻找
int son; //树根的子女结点的个数
int subnets[MAXV]; //记录删除i点后的连通分量
int count; //记录访问


void init(){
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(subnets,0,sizeof(subnets));
son=0;
count=1;
dfn[1]=low[1]=1; //对根节点初始化
vis[1]=1;
}


void dfs(int x){ //tajrjan
for(int i=1;i<=rcnt;i++){
if(map[x][i]){
if(!vis[i]){
vis[i]=1;
dfn[i]=low[i]=++count;
dfs(i);
low[x]=min(low[x],low[i]);
if(low[i]>=dfn[x]){
if(x!=1) subnets[x]++;
if(x==1) son++;
}
}else low[x]=min(low[x],dfn[i]);
}
}
}


void work(){
init();
dfs(1);
}
void Output(){
int flag=0;
if (son>1) subnets[1]=son-1; //到时候输出要+1
for(int i=1;i<=rcnt;i++){
if(subnets[i]){
printf("  SPF node %d leaves %d subnets\n",i,subnets[i]+1);
flag=1;
}
}
if(!flag) printf("  No SPF nodes\n");
printf("\n");
}


void cntv(int x){
rcnt=(rcnt>x?rcnt:x);
}


int main(){
// freopen("d:\\test.in","r",stdin);
int a,b,Case=0;
while(scanf("%d",&a) && a){
memset(map,0,sizeof(map));
scanf("%d",&b);
map[a][b]=map[b][a]=1;
rcnt=0;
cntv(a),cntv(b);
while(scanf("%d",&a) && a){
scanf("%d",&b);
map[a][b]=map[b][a]=1;
cntv(a),cntv(b);
}
work();
printf("Network #%d\n",++Case);
Output();
}
return 0;
}