uva610 Street Directions(无向图桥的应用)

题意

有一张道路交通图，每条边都是双向的，且连通。现在想尽量把一些多的道路改成单向的，且任意两点是可达的，输出改造后的道路图。

分析

同一个双连通分量里面的边都是可以通过重定向使得相互可达，可以改成单向边。桥边就只能是双向的了。

const int maxn = 1010;int dfn[maxn], low[maxn], depth;int belong[maxn], block;stack<int> st;bool in[maxn];vector<ii> bridge;vector<int> G[maxn];bool vis[maxn][maxn];void tarjan(int u,int fa) {    dfn[u] = low[u] = ++depth;    st.push(u);    in[u] = true;    bool had = false;    for (int i = 0;i < G[u].size();++i) {        int v = G[u][i];        if (v == fa && !had) {had = true;continue;}        if (dfn[v] == -1) {            tarjan(v, u);            low[u] = min(low[u], low[v]);            if (dfn[u] < low[v]) {                bridge.push_back(ii(u, v));                bridge.push_back(ii(v, u));                vis[u][v] = vis[v][u] = true;            }else {                bridge.push_back(ii(u, v));                vis[u][v] = vis[v][u] = true;            }        }else if (in[v]) {            low[u] = min(low[u], dfn[v]);            if (!vis[u][v]) {                bridge.push_back(ii(u, v));                vis[u][v] = vis[v][u] = true;            }        }    }}int main(int argc, const char * argv[]){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    // clock_t _ = clock();    int n, m;    while(~scanf("%d%d", &n, &m) && n + m) {        if (n == 0 && m == 0) break;        for (int i = 0;i <= n + 1;++i)            G[i].clear();        bridge.clear();        int u, v;        while(m--) {            scanf("%d%d", &u, &v);            G[u].push_back(v);            G[v].push_back(u);        }        memset(in, false,sizeof in);        memset(dfn, -1,sizeof dfn);        memset(vis, false,sizeof vis);        depth = 0;        tarjan(1, -1);        printf("%d\n\n", ++nCase);        sort(bridge.begin(), bridge.end());        for (int i = 0;i < bridge.size();++i) {            printf("%d %d\n", bridge[i].first, bridge[i].second);        }        puts("#");    }    // printf("\nTime cost: %.2fs\n", 1.0 * (clock() - _) / CLOCKS_PER_SEC);    return 0;}