详情请参考HDU2767 完全相同的题目
问加多少边能使这个图变为一个强连通图。缩点后判断入度和出度的个数，求最大值即可
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#include <vector>
#include <stack>
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;
const int N = 20010;
stack<int> S;
vector<int> G[N];
int dfn[N], low[N], sccno[N], tclock, scccnt;
int id[N], od[N];
int t, n, m, x, y;
void tarjan(int u) {
dfn[u] = low[u] = ++tclock;
S.push(u);
int sz = G[u].size();
repf (i, 0, sz - 1) {
int v = G[u][i];
if (!dfn[v]) { // v not visited
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (!sccno[v]) { // v not belong to scc, so it was in the stack
low[u] = min(low[u], dfn[v]);
}
}
if (low[u] == dfn[u]) {
scccnt++;
int v = -1;
while (v != u) {
v = S.top();
S.pop();
sccno[v] = scccnt;
}
}
}
void find_scc() {
tclock = scccnt = 0;
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(sccno, 0, sizeof(sccno));
repf (i, 1, n) {
if (!dfn[i]) {
tarjan(i);
}
}
}
void read() {
scanf("%d%d", &n, &m);
repf (i, 0, n)
G[i].clear();
while (m--) {
scanf("%d%d", &x, &y);
G[x].push_back(y);
}
}
int solve() {
if (scccnt == 1)
return 0;
memset(id, 0, sizeof(id));
memset(od, 0, sizeof(od));
repf (u, 1, n) {
int sz = G[u].size();
repf (i, 0, sz - 1) {
int v = G[u][i];
if (sccno[u] != sccno[v]) {
id[sccno[v]]++;
od[sccno[u]]++;
}
}
}
int idnum = 0, odnum = 0;
repf (i, 1, scccnt) {
idnum += (id[i] == 0);
odnum += (od[i] == 0);
}
return max(idnum, odnum);
}
int main() {
scanf("%d", &t);
while (t--) {
read();
find_scc();
printf("%d\n", solve());
}
return 0;
}