dfn[u]: 表示节点u的搜索优先级
low[u]: 表示节点u,通过其本身或其子节点能到达的最小有搜索优先级

low[u] = Min{
    1. dfn[u]         其本身搜索优先级
    2. Min{ low[v] }  其子节点v能到达的最小优先级
    3. Min( dfn[v] )  其通过回边(u,v),其中v为u的祖先节点,的优先级
}

 

一 无向图

1. 割点

又名关键点,若删除该点与其发出的边.则整个图不连通.

当前顶点u是一个关键点的充要条件是:

1. 若顶点U是根,则其必定包含两个以上的子节点. (因为若只有一个.删除了U之后,图仍然连通)

2. 若顶点U不是根, 则当 dfn[u] <= low[v] , (其中V是U的子孙节点). 因为V无法通过其本身或子孙到达U或者U更高级的点.所以删除U后,图不连通.

 

poj 1523 SPF

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 1010;

int edge[N][N];
int n, son;

int subnet[N], dfn[N], low[N], tmpdfn;
bool vis[N];

void init(){
    tmpdfn = 1; son = 0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(subnet,0,sizeof(subnet));
    low[1] = dfn[1] = 1; vis[1] = true;
}
void dfs(int u){
   // printf("u = %d\n", u );
    for(int v = 1; v <= n; v++)
    {
        if( edge[u][v] ){
            if( !vis[v] ){
                vis[v] = true;
                dfn[v] = low[v] = ++tmpdfn;
                dfs( v );
                low[u] = min( low[u], low[v] );
                if( low[v] >= dfn[u] ){
                    if( u != 1 ) subnet[u]++;
                    else son++;
                }
            }
            else low[u] = min( low[u], dfn[v] );
        }
    }
}

int main(){
  //  freopen("1.in","r",stdin);
    int Case = 0;
    while( 1 ){
        int u, v; n = 0;
        scanf("%d", &u);
        if( u == 0 ) break;
        memset(edge,0,sizeof(edge));
        scanf("%d",&v);
        if( u > n ) n = u;
        if( v > n ) n = v;
        edge[u][v] = edge[v][u] = 1;
        while(1){
            scanf("%d",&u);
            if( u == 0 ) break;
            scanf("%d",&v);
            if(u > n) n = u;
            if(v > n) n = v;
            edge[u][v] = edge[v][u] = 1;
        }
        if( Case ) puts("");
        printf("Network #%d\n", ++Case);
        init();
        dfs(1);
        if( son > 1 ) subnet[1] = son-1;
        bool find = false;
        for(int i = 1; i <= n; i++)
        {
            if( subnet[i] ){
                find = true;
                printf("  SPF node %d leaves %d subnets\n", i, subnet[i]+1);
            }
        }
        if( !find ) printf("  No SPF nodes\n");
    }
    return 0;
}
View Code

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