LightOJ - 1034 Hit the Light Switches(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：2024/06/11 01:16

题目大意：有N盏灯，灯满足u, v要求，表示u亮了，v也就亮了
问需要开几盏灯，才能让所有的灯都亮了

解题思路：强连通块内的灯能相互作用。求出强连通块，然后缩点连边，找出入度为0的点，入度为0的点，就是要按的灯的数量

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 10010;const int M = 100010;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];int n, m, scc_cnt, dfs_clock, top, cas = 1;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;    }}void dfs(int u, int fa) {    lowlink[u] = pre[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(pre[v], lowlink[u]);        }    }    int x;    if (lowlink[u] == pre[u]) {        scc_cnt++;        while (1) {            x = Stack[top--];            sccno[x] = scc_cnt;            if (x == u) break;        }    }}void solve() {    memset(pre, 0, sizeof(pre));    memset(sccno, 0, sizeof(sccno));    dfs_clock = top = scc_cnt = 0;    for (int i = 1; i <= n; i++) {        if (!pre[i]) dfs(i, -1);    }    for (int i = 1; i <= scc_cnt; i++)        in[i] = 0;    for (int i = 0; i < m; i++) {        int u = sccno[E[i].u];        int v = sccno[E[i].v];        if (u == v) continue;        in[v]++;    }    int ans = 0;    for (int i = 1; i <= scc_cnt; i++)        if (!in[i]) ans++;    printf("Case %d: %d\n", cas++, ans);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}

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