LightOJ - 1034 Hit the Light Switches(强连通)

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
, sccno
, pre
, lowlink
, Stack
, in
;
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;
}