POJ - 1438 One-way Traffic(混合图改有向图)

题目大意：给出一张混合图，要求你改变尽量多的双向边，使得改变后的图还是强连通的

解题思路：这题和poj-1515类似，只不过这题是混合题，大体思路还是差不多的，在dfs的时候记录一下桥和使用的是哪些边即可

#include <cstdio>
#include <cstring>

#define min(a,b)((a) < (b) ? (a) : (b))
#define N 2010
#define M 4000010

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

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

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

    for (int i = head[u]; i != -1; i = E[i].next) {
        int v = E[i].to;
        if (E[i].flag != -1) continue;
        if (E[i].dir == 0) continue;
        E[i].flag = 1;
        E[i^1].flag = 0;

        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]);
        }
    }

}

void solve() {
    memset(pre, 0, sizeof(pre));
    dfs_clock = 0;
    for (int i = 1; i <= n; i++)
        if (!pre[i])
            dfs(i, -1);
    for (int i = 0; i < tot; i += 2) {
        if (E[i].dir == 2 && E[i].flag == 1 && E[i^1].flag == 1) {
            printf("%d %d 2\n", E[i].from, E[i].to);
        }
        else if (E[i].dir == 2 && E[i].flag == 1 && E[i^1].flag == 0) {
            printf("%d %d 1\n", E[i].from, E[i].to);
        }
        else if (E[i].dir == 2 && E[i].flag == 0 && E[i ^ 1].flag == 1) {
            printf("%d %d 1\n", E[i^1].from, E[i^1].to);
        }
    }
}

void AddEdge(int u, int v, int dir) {
    E[tot].from = u;
    E[tot].to = v;
    E[tot].dir = dir;
    E[tot].flag = -1;
    E[tot].next = head[u];
    head[u] = tot++;
}

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

    int a, b, c;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &a, &b, &c);
        if (c == 2) {
            AddEdge(a, b, c);
            AddEdge(b, a, c);
        }
        else {
            AddEdge(a, b, c);
            AddEdge(a, b, 0);
        }
    }
}

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

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