HDU - 3836 Equivalent Sets (强连通分量+DAG)

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题目大意：给出N个点，M条边，要求你添加最少的边，使得这个图变成强连通分量

解题思路：先找出所有的强连通分量和桥，将强连通分量缩点，桥作为连线，就形成了DAG了

这题被坑了。用了G＋＋交的，结果一直RE，用C＋＋一发就过了。。。

#include <cstdio>#include <cstring>#define N 20010#define M 100010#define min(a,b) ((a) > (b)? (b): (a))#define max(a,b) ((a) > (b)? (a): (b))struct Edge{    int from, to, next;}E[M];int head[N], sccno[N], pre[N], lowlink[N], stack[N], in[N], out[N];int n, m, tot, dfs_clock, top, scc_cnt;void AddEdge(int from, int to) {    E[tot].from = from;    E[tot].to = to;    E[tot].next = head[from];    head[from] = 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);    }}void dfs(int u) {    pre[u] = lowlink[u] = ++dfs_clock;    stack[++top] = u;    for (int i = head[u]; i != -1; i = E[i].next) {        int v = E[i].to;        if (!pre[v]) {            dfs(v);            lowlink[u] = min(lowlink[u], lowlink[v]);        }        else if (!sccno[v]) {            lowlink[u] = min(lowlink[u], pre[v]);        }    }    int x;    if (pre[u] == lowlink[u]) {        scc_cnt++;        while (1) {            x = stack[top--];            sccno[x] = scc_cnt;            if (x == u)                break;        }    }}void solve() {    memset(sccno, 0, sizeof(sccno));    memset(pre, 0, sizeof(pre));    dfs_clock = top = scc_cnt = 0;    for (int i = 1; i <= n; i++)        if (!pre[i])             dfs(i);    if (scc_cnt <= 1) {        printf("0\n");        return ;    }    for (int i = 1; i <= scc_cnt; i++)         in[i] = out[i] = 1;    for (int i = 0; i < tot; i++) {        int u = E[i].from, v = E[i].to;        if (sccno[u] != sccno[v]) {            out[sccno[u]] = in[sccno[v]] = 0;        }    }    int a = 0, b = 0;    for (int i = 1; i <= scc_cnt; i++) {        if (out[i]) a++;        if (in[i]) b++;    }    printf("%d\n", max(a, b));}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        init();        solve();    }    return 0;}

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