LightOJ - 1210 Efficient Traffic System(强连通)

来源：互联网发布：网络维护与故障排解编辑：程序博客网时间：2024/06/06 12:29

题目大意：给出一张有向图，问需要添加多少条边，才能使该有向图变成强连通图

解题思路：强连通，缩点，算出每点的入度和出度，找出max（入度为0，出度为0），这个就是答案
解释一下为什么，如果要强连通的话，就得让所有点都有入度和出度，所以添加的边的数量就是max(入度为0，出度为0)

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 20010;const int M = 50010;struct Edge{    int u, v, next;    Edge() {}    Edge(int u, int v, int next): u(u), v(v), next(next) {}}E[M];int head[N], sccno[N], pre[N], lowlink[N], Stack[N], in[N], out[N];int n, m, scc_cnt, dfs_clock, top, cas = 1;void dfs(int u, int fa) {    pre[u] = lowlink[u] = ++dfs_clock;    Stack[++top] = u;    for (int i = head[u]; ~i; i = E[i].next) {        int v = E[i].v;        if (!pre[v]) {            dfs(v, u);            lowlink[u] = min(lowlink[u], lowlink[v]);        }        else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]);    }    if (pre[u] == lowlink[u]) {        scc_cnt++;        int x;         while (1) {            x = Stack[top--];            sccno[x] = scc_cnt;            if (x == u) break;        }    }}void solve() {    if (n <= 1) {        printf("Case %d: %d\n", cas++, 0);        return ;    }    memset(pre, 0, sizeof(pre));    memset(sccno, 0, sizeof(sccno));    scc_cnt = dfs_clock = top = 0;    for (int i = 1; i <= n; i++)        if (!pre[i]) dfs(i, -1);    if (scc_cnt == 1) {        printf("Case %d: %d\n", cas++, 0);        return ;    }    memset(in, 0, sizeof(in));    memset(out, 0, sizeof(out));    for (int i = 0; i < m; i++) {        int u = sccno[E[i].u];        int v = sccno[E[i].v];        if (u == v) continue;        out[u] = 1;        in[v] = 1;    }    int in_cnt = 0, out_cnt = 0;    for (int i = 1; i <= scc_cnt; i++) {        if (!in[i]) in_cnt++;        if (!out[i]) out_cnt++;    }    printf("Case %d: %d\n", cas++, max(in_cnt, out_cnt));}void init() {    memset(head, -1, sizeof(head));    scanf("%d%d", &n, &m);    int u, v;    for (int i = 0; i < m; i++) {        scanf("%d%d", &u, &v);        E[i] = Edge(u, v, head[u]);        head[u] = i;    }}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}

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