POJ 1515 Street Directions 边双连通分量 + dfs

题目：http://poj.org/problem?id=1515

题意：给定一个无向连通图，对图中的尽量多的边定为单向边，使之成为一个强连通图，对于无法定成单向边的，就定为双向边

思路：显然，对图进行边双连通分量分解，其中桥一定定为双向边，否则必然不是强连通图，对于其他边，因为是连通分量内的边，必然成环，按dfs遍历输出就好，必定构成强连通图

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 1010;
struct edge
{
int to, next;
} G[N*100];
int dfn
, low
, head
, st
, dcc
;
int index, num, cnt, top;
int n, m, x = 0;
bool vis
, visnode
, visedge[N*100];
void init()
{
memset(dfn, -1, sizeof dfn);
memset(head, -1, sizeof head);
memset(vis, 0, sizeof vis);
memset(visnode, 0, sizeof visnode);/*标记顶点*/
memset(visedge, 0, sizeof visedge);/*标记边*/
index = num = cnt = top = 0;
}
void add_edge(int v, int u)
{
G[cnt].to = u;
G[cnt].next = head[v];
head[v] = cnt++;
}
bool judge(int v, int fa)
{
int tmp = 0;
for(int i = head[v]; i != -1; i = G[i].next)
if(G[i].to == fa) tmp++;
if(tmp >= 2) return true;
else return false;
}
void tarjan(int v, int fa)
{
dfn[v] = low[v] = index++;
vis[v] = true;
st[top++] = v;
bool f = judge(v, fa);
int u, t;
for(int i = head[v]; i != -1; i = G[i].next)
{
u = G[i].to;
if(u == fa && !f) continue;
if(dfn[u] == -1)
{
tarjan(u, v);
low[v] = min(low[v], low[u]);
}
else if(vis[u])
low[v] = min(low[v], dfn[u]);
}
if(dfn[v] == low[v]) /*v不能连到祖先，必为连通分量的根*/
{
num++;
do
{
t = st[--top];
dcc[t] = num;
vis[t] = false;
}
while(t != v);
}
}
void dfs(int v, int fa)
{
visnode[v] = true;
for(int i = head[v]; i != -1; i = G[i].next)
{
int u = G[i].to;
if(u == fa) continue;
if(dcc[v] == dcc[u] && visedge[i] == false) /*同一个边连通分量内，定为单向边*/
{
printf("%d %d\n", v, u);
}
else if(dcc[v] != dcc[u] && visedge[i] == false) /*边（v,u）为桥，双向边*/
{
printf("%d %d\n", v, u);
printf("%d %d\n", u, v);
}
visedge[i] = true; /*无向图，所以标记两次*/
visedge[i^1] = true;
if(! visnode[u])
dfs(u, v);
}
}
void slove()
{
printf("%d\n\n", ++x);
tarjan(1, -1);
dfs(1, -1);
printf("#\n");
}
int main ()
{
int a, b;
while(scanf("%d%d", &n, &m), n || m)
{
init();
for(int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
add_edge(a, b);
add_edge(b, a);
}
slove();
}

return 0;
}