UVa610 - Street Directions(桥的问题)

Street Directions

According to the Automobile Collision Monitor (ACM), most fatal trafficaccidents occur ontwo-way streets. In order to reduce the number of fatalities caused bytraffic accidents, themayor wants to convert as many streets as possible into one-way streets.You have been hiredto perform this conversion, so that from each intersection, it ispossible for a motorist to driveto all the other intersections following some route.

You will be given a list of streets (all two-way) of the city.Each street connects two intersections,and does not go through an intersection. At most four streetsmeet at each intersection, andthere is at most one street connecting any pair of intersections.It is possible for an intersectionto be the end point of only one street. You may assume that itis 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 number of cases. The first line of each casecontains two integers n andm. The number of intersections is n (), and thenumber of streets is m. Thenext m lines contain the intersections incident to each of the mstreets. The intersections arenumbered from 1 to n, and each street is listed once. If thepair is present, will not bepresent. End of input is indicated by n = m = 0.

Output 

For each case, print the case number (starting from 1) followedby a blank line. Next, printon separate lines each street as the pair to indicate thatthe street has been assigned thedirection going from intersection i to intersection j. For astreet that cannot be converted intoa one-way street, print both and on two different lines.The list of streets can be printedin any order. Terminate each case with a line containing asingle `#' 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
#

桥的查找问题

#include <cstdio>
#include <cstring>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

const int N = 1010;

vector<int> g[N];
bool vis[N][N];
vector< pair<int, int> > ans;
int n, m;
int low[N], dfn[N], dfs_clock;
int t = 1;

bool input();
void solve();
void dfs(int u, int father);

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("d:\\OJ\\uva_in.txt", "r", stdin);
	#endif
	
	while (input()) {
		solve();
	}
	return 0;
}

bool input()
{
	scanf("%d%d", &n, &m);
	if (n == 0 && m == 0) return false;
	
	for (int i = 0; i < N; i++) g[i].clear();
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	return true;
}

void dfs(int u, int father)
{
	dfn[u] = low[u] = ++dfs_clock;
	
	for (size_t i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == father) continue;
		if (dfn[v] == -1) {
			dfs(v, u);
			low[u] = min(low[v], low[u]);
			if (low[v] > dfn[u]) {
				ans.push_back(make_pair(u, v));
				ans.push_back(make_pair(v, u));
				vis[u][v] = vis[v][u] = true;
			} else {
				ans.push_back(make_pair(u, v));
				vis[u][v] = vis[v][u] = true;
			}
		} else {
			low[u] = min(low[u], dfn[v]);
			if (!vis[u][v]) {
				ans.push_back(make_pair(u, v));
				vis[u][v] = vis[v][u] = true;
			}
		}
	}
}

void solve()
{
	memset(dfn, -1, sizeof(dfn));
	memset(vis, false, sizeof(vis));
	dfs_clock = 0;
	ans.clear();
	
	dfs(1, 1);
	
	printf("%d\n\n", t++);
	for (size_t i = 0; i < ans.size(); i++) {
		printf("%d %d\n", ans[i].first, ans[i].second);
	}
	printf("#\n");
}