poj1523--C - SPF（连通分量，求割点）

C - SPF

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

 

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 25 43 13 23 43 501 22 33 44 55 101 22 33 44 66 32 55 100

Sample Output

Network #1  SPF node 3 leaves 2 subnetsNetwork #2  No SPF nodesNetwork #3  SPF node 2 leaves 2 subnets  SPF node 3 leaves 2 subnets

给出无向边，输入到0结束，问割点是谁，并且求出割点将图分成几个连通分量

割点的求法就是tarjan，在求出割点后用并查集找出将图分成几块，注意对于根，如果根只有一个子树，那么根不是割点，但用tarjan判断也会判断成割点，所以要特判根节点

所以，如果割点找出的分块数是1，不用输出。

输出是坑，要求每个空两个格，一个案例空一行

 

 

#include <cstdio>#include <cstring>#include <stack>#include <algorithm>using namespace std;#define maxn 1200struct node{    int u , v ;    int next ;} edge[1000000] ;int head[maxn] , cnt , vis[1000000] ;int dnf[maxn] , low[maxn] , time ;int ans[maxn] ;int c[maxn] , n ;stack <int> sta;void init(){    memset(head,-1,sizeof(head));    memset(vis,0,sizeof(vis));    memset(dnf,0,sizeof(dnf));    memset(low,0,sizeof(low));    memset(ans,0,sizeof(ans));    cnt = time = n = 0 ;}void add(int u,int v){    edge[cnt].u = u ;    edge[cnt].v = v ;    edge[cnt].next = head[u] ;    head[u] = cnt++ ;    edge[cnt].u = v ;    edge[cnt].v = u ;    edge[cnt].next = head[v] ;    head[v] = cnt++ ;}int find1(int x){    int r , k , l ;    r = x ;    while( r != c[r] )        r = c[r] ;    k = x ;    while( k != r )    {        l = c[k] ; c[k] = r ; k = l ;    }    return r ;}int f(int s){    int i , j , u , v , num[maxn] , sum = 0 ;    memset(num,0,sizeof(num));    for(i = 1 ; i <= n ; i++)        c[i] = i ;    for(i = 1 ; i <= n ; i++)    {        for(j = head[i] ; j != -1 ; j = edge[j].next)        {            if( edge[j].u == s || edge[j].v == s )                continue ;            u = find1( edge[j].u ) ;            v = find1( edge[j].v ) ;            if(u != v)                c[u] = v ;        }    }    for(i = 1 ; i <= n ; i++)    {        if( head[i] != -1 && i != s )        {            num[ find1(i) ]++ ;        }    }    for(i = 1 ; i <= n ; i++)        if( num[i] )            sum++ ;    return sum ;}void tarjan(int u){    dnf[u] = low[u] = ++time ;    int v, i ;    for(i = head[u] ; i != -1 ; i = edge[i].next)    {        if( vis[i] ) continue ;        vis[i] = vis[i^1] = 1 ;        v = edge[i].v ;        if( !dnf[v] )        {            sta.push(i);            tarjan(v);            low[u] = min( low[u],low[v] );            if( low[v] >= dnf[u] && !ans[u] )            {                ans[u] = f(u);            }        }        else if( dnf[v] < dnf[u] )            low[u] = min( low[u],dnf[v] );    }}int main(){    int u , v , temp = 0 , i;    while(scanf("%d", &u) && u)    {        temp++ ;        init();        scanf("%d", &v);        add(u,v);        n = max(n,u);        n = max(n,v);        while(scanf("%d", &u) && u)        {            scanf("%d", &v);            add(u,v);            n = max(n,u);            n = max(n,v);        }        tarjan(1);        int flag = 0 ;        printf("Network #%d\n", temp);        for(i = 1 ; i <= n ; i++)            if( ans[i] > 1 )            {                printf("  SPF node %d leaves %d subnets\n", i, ans[i]);                flag = 1 ;            }        if( !flag )            printf("  No SPF nodes\n\n");        else            printf("\n");    }    return 0;}