UVA 796 - Critical Links （求桥）

Critical Links 

In a computer network a link L, which interconnects two servers, is considered critical if there are at least two servers A and B such that all network interconnection paths between A and B pass through L. Removing a critical link generates two disjoint sub-networks such that any two servers of a sub-network are interconnected. For example, the network shown in figure 1 has three critical links that are marked bold: 0 -1, 3 - 4 and 6 - 7.

Figure 1: Critical links

It is known that:

1.

the connection links are bi-directional;

2.

a server is not directly connected to itself;

3.

two servers are interconnected if they are directly connected or if they are interconnected with the same server;

4.

the network can have stand-alone sub-networks.

Write a program that finds all critical links of a given computer network.

Input

The program reads sets of data from a text file. Each data set specifies the structure of a network and has the format:

...

The first line contains a positive integer (possibly 0) which is the number of network servers. The next  lines, one for each server in the network, are randomly ordered and show the way servers are connected. The line corresponding to server_k, , specifies the number of direct connections of server_k and the servers which are directly connected to server_k. Servers are represented by integers from 0 to . Input data are correct. The first data set from sample input below corresponds to the network in figure 1, while the second data set specifies an empty network.

Output

The result of the program is on standard output. For each data set the program prints the number of critical links and the critical links, one link per line, starting from the beginning of the line, as shown in the sample output below. The links are listed in ascending order according to their first element. The output for the data set is followed by an empty line.

Sample Input

8
0 (1) 1
1 (3) 2 0 3
2 (2) 1 3
3 (3) 1 2 4
4 (1) 3
7 (1) 6
6 (1) 7
5 (0)

0

Sample Output

3 critical links
0 - 1
3 - 4
6 - 7

0 critical links

模板题

需要按照顺序输出桥

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
/*
*  求 无向图的割点和桥
*  可以找出割点和桥，求删掉每个点后增加的连通块。
*  需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
const int MAXN = ;
const int MAXM = ;
struct Edge
{
    int to,next;
    bool cut;//是否为桥的标记
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int bridge;

void addedge(int u,int v)
{
    edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;
    head[u] = tot++;
}

void Tarjan(int u,int pre)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    int son = ;
    for(int i = head[u];i != -;i = edge[i].next)
    {
        v = edge[i].to;
        if(v == pre)continue;
        if( !DFN[v] )
        {
            son++;
            Tarjan(v,u);
            if(Low[u] > Low[v])Low[u] = Low[v];
            //桥
            //一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
            if(Low[v] > DFN[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^].cut = true;
            }
            //割点
            //一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。
            //(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，
            //即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)
            if(u != pre && Low[v] >= DFN[u])//不是树根
            {
                cut[u] = true;
                add_block[u]++;
            }
        }
        else if( Low[u] > DFN[v])
             Low[u] = DFN[v];
    }
    //树根，分支数大于1
    if(u == pre && son > )cut[u] = true;
    if(u == pre)add_block[u] = son - ;
    Instack[u] = false;
    top--;
}

void solve(int N)
{
    memset(DFN,,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(add_block,,sizeof(add_block));
    memset(cut,false,sizeof(cut));
    Index = top = ;
    bridge = ;
    for(int i = ;i <= N;i++)
        if( !DFN[i] )
            Tarjan(i,i);
    printf("%d critical links\n",bridge);
    vector<pair<int,int> >ans;
    for(int u = ;u <= N;u++)
        for(int i = head[u];i != -;i = edge[i].next)
            if(edge[i].cut && edge[i].to > u)
            {
                ans.push_back(make_pair(u,edge[i].to));
            }
    sort(ans.begin(),ans.end());
    //按顺序输出桥
    for(int i = ;i < ans.size();i++)
        printf("%d - %d\n",ans[i].first-,ans[i].second-);
    printf("\n");
}
void init()
{
    tot = ;
    memset(head,-,sizeof(head));
}
//处理重边
map<int,int>mapit;
inline bool isHash(int u,int v)
{
    if(mapit[u*MAXN+v])return true;
    if(mapit[v*MAXN+u])return true;
    mapit[u*MAXN+v] = mapit[v*MAXN+u] = ;
    return false;
}
int main()
{
    int n;
    while(scanf("%d",&n) == )
    {
        init();
        int u;
        int k;
        int v;
        //mapit.clear();
        for(int i = ;i <= n;i++)
        {
            scanf("%d (%d)",&u,&k);
            u++;
            //这样加边，要保证正边和反边是相邻的，建无向图
            while(k--)
            {
                scanf("%d",&v);
                v++;
                if(v <= u)continue;
                //if(isHash(u,v))continue;
                addedge(u,v);
                addedge(v,u);
            }
        }
        solve(n);
    }
    return ;
}