lightoj1210Efficient Traffic System

思路：给定一个有向图，问至少需要添加几条有向边可以使得图成为一个强连通图。

显然是需要缩点的，因为一个环上的点是可以互达的，可以看成一个点，如果图本

身就是强连通的输出0，因为不需要添加。否则就看缩点后每个点的入度与出度。

没有出的，要添加，没有入的要添加，，，这里就去其最大值（因为没有入的可以

和没有出的点之间加边，剩下的就要和其他任何点之间建边了）。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define MEM(x,y) memset(x, y,sizeof x)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 20010;
int head[maxn], to[maxn << 2], nxt[maxn << 2];
int dfn[maxn], low[maxn], belong[maxn];
bool vis[maxn];
int Time;
int scc;
int n ,m;
stack<int> st;
int tol;
void add(int u,int v){
to[tol] = v;
nxt[tol] = head[u];
head[u] = tol++;
}
void Tarjan(int u){
dfn[u] = low[u] = Time++;
vis[u] = true;
st.push(u);
for (int i = head[u];i != -1;i = nxt[i]){
int v = to[i];
if (dfn[v] == -1){
Tarjan(v);
low[u] = min(low[u], low[v]);
}else if (vis[v] && dfn[v] < low[u]){
low[u] = dfn[v];
}
}
if (dfn[u] == low[u]){
scc++;
while(true){
int v = st.top();
st.pop();
vis[v] = false;
belong[v] = scc;
if (v == u) break;
}
}
}
int in[maxn], out[maxn];
void dfs(int u){
dfn[u] = 1;
for (int i = head[u];i != -1;i = nxt[i]){
int v = to[i];
int a = belong[u];
int b = belong[v];
if (a != b){
in[b]++;
out[a]++;
}
if (dfn[v] == -1) dfs(v);
}
}
int solve(){
if (scc == 1) return 0;
int a = 0,b = 0;
for (int i = 1;i <= scc;++i){
if (!in[i]) a++;
if (!out[i]) b++;
}
return max(a,b);
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int icase = 0, t;
cin >> t;
while(t--){
scanf("%d%d",&n,&m);
tol = scc = 0;
Time = 0;
MEM(belong, -1);
MEM(head, -1);
MEM(vis, false);
MEM(dfn, -1);
for (int i = 0, a, b;i < m;++i){
scanf("%d%d",&a, &b);
add(a,b);
}
for (int i = 1;i <= n;++i)
if (dfn[i] == -1) Tarjan(i);
// cout << "scc = " << scc << endl;
MEM(dfn, -1);
MEM(in, 0);
MEM(out, 0);
for (int i = 1;i <= n;++i)
if (dfn[i] == -1) dfs(i);
printf("Case %d: %d\n", ++icase, solve());
}
return 0;
}