ACM: 无向图的强连通分量 图论题 p…

                                     StreetDirections

 

Description

According to the AutomobileCollision Monitor (ACM), most fatal traffic accidents occur ontwo-way streets. In order to reduce the number of fatalities causedby traffic accidents, the mayor wants to convert as many streets aspossible into one-way streets. You have been hired to perform thisconversion, so that from each intersection, it is possible for amotorist to drive to all the other intersections following someroute.

You will be given a list of streets (all two-way) of the city. Eachstreet connects two intersections, and does not go through anintersection. At most four streets meet at each intersection, andthere is at most one street connecting any pair of intersections.It is possible for an intersection to be the end point of only onestreet. You may assume that it is possible for a motorist to drivefrom each destination to any other destination when every street isa two-way street.

Input

The input consists of a numberof cases. The first line of each case contains two integers n andm. The number of intersections is n (2 <= n<= 1000), and the number of streets is m. The next mlines contain the intersections incident to each of the m streets.The intersections are numbered from 1 to n, and each street islisted once. If the pair i j is present, j i will not be present.End of input is indicated by n = m = 0.

Output

For each case, print the casenumber (starting from 1) followed by a blank line. Next, print onseparate lines each street as the pair i j to indicate that thestreet has been assigned the direction going from intersection i tointersection j. For a street that cannot be converted into aone-way street, print both i j and j i on two different lines. Thelist of streets can be printed in any order. Terminate each casewith a line containing a single `#' character.

Note: There may be many possible direction assignments satisfyingthe requirements. Any such assignment is acceptable.

Sample Input

7 10
1 2
1 3
2 4
3 4
4 5
4 6
5 7
6 7
2 5
3 6
7 9
1 2
1 3
1 4
2 4
3 4
4 5
5 6
5 7
7 6
0 0

Sample Output

1

1 2
2 4
3 1
3 6
4 3
5 2
5 4
6 4
6 7
7 5
#
2

1 2
2 4
3 1
4 1
4 3
4 5
5 4
5 6
6 7
7 5
#

 

题意: 给出一个无向图, 去掉一些边后仍然是连通. 输出不可以转化为, 单向通道的边.

     例如(i,j) 只输出i j 或则 j i即可.

解题思路:

         1. 求无向图的边连通分量, 在同一个连通分量里面, 保留单向的边即可

           即: 构成一个强连通分量. 而对于不同的连通分量, 期间的边必须是双向边,

               才能够保证点与点之间相互连通.

         2. tarjan算法学习。。。

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 1005
const int INF = (1<<29);

struct node
{
 int v;
 int next;
 bool vis;
}edges[MAX*10];

int n, m;
int first[MAX], num;
int bcc[MAX*10], bcnt, deep[MAX*10], low[MAX*10];
int stack[MAX*10], top;
bool instack[MAX*10], vis[MAX*10];
int start, fa_start;

void init()
{
 for(int i = 0; i <= n+2;++i)
 {
  first[i] = -1; // 初始化邻接表头
  deep[i] = low[i] = 0; //初始化深度和从i出发经过的顶点形成路径,
        // 所能到达的最小深度"优先"编号
  bcc[i] = 0;
  vis[i] = instack[i] =false;
 }
 num = top = bcnt = 0;
}

inline void add(int u,int v)
{
 edges[num].v = v;
 edges[num].vis = false;
 edges[num].next = first[u];
 first[u] = num++;
}

void read_graph()
{
 int u, v;
 int i;
 init();
 for(i = 0; i < m; ++i)
 {
  scanf("%d%d",&u,&v);
  add(u,v);
  add(v,u);
 }
 start = 1; //初始化起始点编号.
 fa_start = 0; //初始化起始点的父亲节点编号
}

void tarjan(int u,int fa,int dfsindex)
{
 deep[u] = low[u] = dfsindex++;
 instack[u] = true;
 stack[top++] = u;
 for(int e = first[u]; e != -1; e =edges[e].next)
 {
  int v = edges[e].v;
  if(v != fa)
  {
   if( !deep[v])
   {
    tarjan(v,u,dfsindex);
    if(low[u]> low[v]) low[u] = low[v];
   }
   elseif(instack[v] && deep[v]< low[u])
   {
    low[u]= deep[v];
   }
  }
 }
 if(deep[u] == low[u])
 {
  bcnt++; //计数器
  int v;
  do
  {
   v =stack[top-1];
   top--;
   instack[v] =0;
   bcc[v] =bcnt;
  }while( v != u );
 }
}

void solve(int u,int fa)
{
 for(int e = first[u]; e != -1; e =edges[e].next)
 {
  int v = edges[e].v;
  if(v != fa)
  {
   if(bcc[u] ==bcc[v] && !edges[e].vis)
   {
    printf("%d%d\n",u,v);
   }
   elseif(bcc[u] != bcc[v] &&!edges[e].vis)
   {
    printf("%d%d\n",u,v);
    printf("%d%d\n",v,u);
   }
   edges[e].vis= true;
   edges[e^1].vis= true;
   if( !vis[v])
   {
    vis[v]= true;
    solve(v,u);
   }
  }
 }
}

int main()
{
 int k = 1;
// freopen("input.txt","r",stdin);
 while(scanf("%d%d",&n,&m) != EOF)
 {
  if(n == 0&& m == 0) break;
  
  printf("%d\n\n",k++);
  read_graph();
  tarjan(start,fa_start,1);//start: 起始点, fa_start: 起始点的父亲, 1:初始化深度为1;
  vis[start] = 1;

  solve(start,fa_start);
  printf("#\n");
 }
 return 0;
}