（beginer）DFS （桥...边-双连通分量） UVA 610 - Street Directions

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Street Directions

According to the Automobile Collision Monitor (ACM), most fatal traffic accidents occur on two-way streets. In order to reduce the number of fatalities caused by traffic accidents, the mayor wants to convert as many streets as possible into one-way streets. You have been hired to perform this conversion, so that from each intersection, it is possible for a motorist to drive to 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 streets meet at each intersection, and there is at most one street connecting any pair of intersections. It is possible for an intersection to be the end point of only one street. You may assume that it is possible for a motorist to drive from each destination to any other destination when every street is a two-way street.

Input

The input consists of a number of cases. The first line of each case contains two integersn and m. The number of intersections is n ( $2 \le? n \le? 1000$ ), and the number of streets ism. The next m lines contain the intersections incident to each of them streets. The intersections are numbered from 1 to n, and each street is listed once. If the pair $i\ j$ is present, $j\ i$ will not be present. End of input is indicated byn = m = 0.

Output

For each case, print the case number (starting from 1) followed by a blank line. Next, print on separate lines each street as the pair $i\ j$ to indicate that the street has been assigned the direction going from intersectioni to intersection j. For a street that cannot be converted into a one-way street, print both $i\ j$ and $j\ i$ on two different lines. The list of streets can be printed in any order. Terminate each case with a line containing a single `#' character.

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

Sample Input

7 101 21 32 43 44 54 65 76 72 53 67 91 21 31 42 43 44 55 65 77 60 0

Sample Output

11 22 43 13 64 35 25 46 46 77 5#21 22 43 14 14 34 55 45 66 77 5#

题意：有很多双向的街道，但是我们想要把尽可能多的街道变成单向的，问要怎么破。

思路：首先我们找出整个图中的桥，因为这些街道一定要是双向的。剩下的街道可以构成若干个边-双连通分量，即每一条边都在至少一个简单环中，那么在剩下的边里面dfs一下，dfs的时候不经过桥，那么dfs走的方向就是我们留下来的方向了，哈哈！！

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<stack>#include<string.h>using namespace std;const int maxn = 1000+10;struct Edge{Edge(int uu,int vv) : u(uu) , v(vv) { }int u;int v;};bool cmp(const Edge & e1 , const Edge & e2){if (e1.u==e2.u) return e1.v < e2.v;return e1.u < e2.u;}vector<int> G[maxn];vector<Edge> ans;int dfs_clock , n , m , pre[maxn];bool vis[maxn][maxn];bool is_bridge[maxn][maxn];void init(){ans.clear();memset(pre,0,sizeof(pre));dfs_clock = 0;for (int i = 1 ; i <= n ; ++i) G[i].clear();}void input(){int u , v;while (m--){scanf("%d%d",&u,&v);is_bridge[u][v] = is_bridge[v][u] = vis[u][v] = vis[v][u] = false;G[u].push_back(v);G[v].push_back(u);}}int dfs(int u,int fa){int lowu = pre[u] = ++dfs_clock;for (int i = 0 ; i < G[u].size() ; ++i){int v = G[u][i];if (v==fa) continue;if (!pre[v]){int lowv = dfs(v,u);lowu = min(lowu,lowv);if (lowv > pre[u]) {is_bridge[u][v] = is_bridge[v][u] = true;ans.push_back(Edge(u,v));ans.push_back(Edge(v,u));}} else if (lowu > pre[v]) lowu = pre[v];}return lowu;}void dfs1(int u,int fa){for (int i = 0 ; i < G[u].size() ; ++i){int v = G[u][i];if (is_bridge[u][v] || vis[u][v] || v==fa) continue;vis[u][v] = vis[v][u] = true;ans.push_back(Edge(u,v));dfs1(v,u);}}void solve(){dfs(1,-1);for (int i = 1 ; i <= n ; ++i) dfs1(i,-1);sort(ans.begin(),ans.end(),cmp);for (int i = 0 ; i < ans.size() ; ++i) printf("%d %d\n",ans[i].u,ans[i].v);}int main(){int k = 0;while (scanf("%d%d",&n,&m),n+m){++k;init();input();printf("%d\n\n",k);solve();printf("#\n");}}

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