POJ - 1515 Street Directions(无向图变有向图)

题目大意：给出一张无向图，任意两点可相互到达，现在要求你将这张无向图变成有向图，且改变之后任意两点还是可相互到达的

解题思路：桥的话肯定是要保留双向的，所以可以在dfs的时候把桥标记出来，顺便在dfs的时候标记一下使用的是哪条边即可

#include <cstdio>
#include <cstring>

#define N 1010
#define M 1000010
#define min(a,b) ((a) < (b) ? (a) : (b))

struct Edge{
    int from, to, next, flag;
}E[M];

int head[N], pre[N], lowlink[N];
int tot, n, m, dfs_clock;

void AddEdge(int u, int v) {
    E[tot].from = u;
    E[tot].to = v;
    E[tot].flag = 0;
    E[tot].next = head[u];
    head[u] = tot++;
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;

    int u, v;
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &u, &v);
        AddEdge(u, v);
        AddEdge(v, u);
    }
}

void dfs(int u, int fa) {
    pre[u] = lowlink[u] = ++dfs_clock;

    int v;
    for (int i = head[u]; i != -1; i = E[i].next) {
        v = E[i].to;
        if (E[i].flag) continue;
        E[i].flag = 1;
        E[i ^ 1].flag = -1;
        if (!pre[v]) {
            dfs(v, u);
            lowlink[u] = min(lowlink[u], lowlink[v]);
            if (lowlink[v] > pre[u])
                E[i ^ 1].flag = 1;
        }
        else if (v != fa)
            lowlink[u] = min(lowlink[u], pre[v]);
    }

}

int cas = 1;
void solve() {
    memset(pre, 0, sizeof(pre));
    dfs_clock = 0;

    for (int i = 1; i <= n; i++)
        if (!pre[i])
            dfs(i, -1);

    printf("%d\n\n", cas++);
    int u, v;
    for (int i = 0; i < tot; i++) {
        u = E[i].from;
        v = E[i].to;
        if (E[i].flag == 1)
            printf("%d %d\n", u, v);
    }
    printf("#\n");
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        init();
        solve();
    }
    return 0;
}

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