给一个无向图,要求变成强连通的有向图,需要保留哪些边。
边的双连通,对于桥保留两条边,其他的只保留一条边。求双连通的过程中记录保留边。
/*********************************************
Problem: 1515 User: G_lory
Memory: 232K Time: 32MS
Language: C++ Result: Accepted
**********************************************/
#include <cstdio>
#include <cstring>
#include <iostream>
#define pk printf("KKK!\n");
using namespace std;
const int N = 1005;
const int M = N * N;
struct Edge {
int from, to, next;
int cut;
} edge[M];
int cnt_edge;
int head[N];
void add_edge(int u, int v)
{
edge[cnt_edge].from = u;
edge[cnt_edge].to = v;
edge[cnt_edge].next = head[u];
edge[cnt_edge].cut = 0;
head[u] = cnt_edge++;
}
int dfn[N]; int idx;
int low[N];
int n, m;
void tarjan(int u, int pre)
{
dfn[u] = low[u] = ++idx;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cut) continue;
edge[i].cut = 1;
edge[i ^ 1].cut = -1;
if (v == pre) continue;
if (!dfn[v])
{
tarjan(v, u);
low[u] = min(low[u], low[v]);
if (dfn[u] < low[v])
edge[i].cut = edge[i ^ 1].cut = 1;
}
else low[u] = min(low[u], dfn[v]);
}
}
void init()
{
idx = cnt_edge = 0;
memset(dfn, 0, sizeof dfn);
memset(head, -1, sizeof head);
}
void solve()
{
for (int i = 0; i < cnt_edge; ++i)
if (edge[i].cut == 1)
printf("%d %d\n", edge[i].from, edge[i].to);
}
int main()
{
//freopen("in.txt", "r", stdin);
int cas = 0;
while (~scanf("%d%d", &n, &m))
{
if (n == 0 && m == 0) break;
printf("%d\n\n", ++cas);
int u, v;
init();
for (int i = 0; i < m; ++i)
{
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
tarjan(1, -1);
solve();
printf("#\n");
}
return 0;
}
poj1515--Street Directions(边的双连通)
原文:http://www.cnblogs.com/wenruo/p/5007998.html