LightOJ - 1417 Forwarding Emails(强连通+dfs)

题目大意：你有一个秘密，但你只能告诉一个。这个人如果有认识的人的话，他就会把你的秘密告诉出去，这样就一传十，十传百了。现在你想要知道，最多能有多少人知道

解题思路：同一个连通分量内的人都是可知的，所以缩点，然后连边，从入度为0的点出发，dfs找出所能到的最远处，在dfs过程中统计一下有多少个人知道

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const int N = 50010;

struct Edge{
int u, v, next;
Edge() {}
Edge(int u, int v, int next): u(u), v(v), next(next) {}
}E
;

int head
, pre
, lowlink
, sccno
, num
, Stack
, in
;
int dfs_clock, scc_cnt, top, n, m, cas = 1;
bool vis
;
vector<int> V
;

void init() {
scanf("%d", &n);
memset(head, -1, sizeof(head));

int u, v;
for (int i = 0; i  < n; i++) {
scanf("%d%d", &u, &v);
E[i] = Edge(u, v, head[u]);
head[u] = i;
}
}

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 = 0;
num[scc_cnt] = 0;
while (1) {
x = Stack[top--];
sccno[x] = scc_cnt;
num[scc_cnt]++;
if (x == u) break;
}
}
}

int dfs2(int u){
int Max = 0;
for (int i = 0; i < V[u].size(); i++) {
int v = V[u][i];
Max = max(Max, dfs2(v));
}
return Max + num[u];
}

void solve() {
memset(pre, 0, sizeof(pre));
memset(sccno, 0, sizeof(sccno));
dfs_clock = scc_cnt = 0;

for (int i = 1; i <= n; i++)
if (!pre[i]) dfs(i, -1);

if (scc_cnt == 1) {
printf("Case %d: %d\n", cas++, 1);
return ;
}

memset(in, 0, sizeof(in));

for (int i = 1; i <= scc_cnt; i++)
V[i].clear();

for (int i = 0; i < n; i++) {
int u = sccno[E[i].u];
int v = sccno[E[i].v];

if (u == v) continue;
in[v]++;
V[u].push_back(v);
}
int Max = 0, pos;
for (int i = 1; i <= scc_cnt; i++)
if (!in[i]) {
int t = dfs2(i);
if (t > Max) {
Max = t;
pos = i;
}
}
for (int i = 1; i <= n; i++)
if (sccno[i] == pos) {
printf("Case %d: %d\n", cas++, i);
return ;
}
}

int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}