POJ 1236 ：http://poj.org/problem?id=1236
参考：https://www.cnblogs.com/TnT2333333/p/6875680.html
题意：
有好多学校，每个学校可以给其他特定的学校发送文件。第一个问题是最少要给几个学校发文件，可以使得全部的学校收到文件。第二个问题是最少要加几条线路，使得随意挑一个学校发文件，也能使得全部的学校收到文件。
思路：
第一个问题，可以用tarjan给图中先缩点，因为强连通的环相互可达。所以只要数出缩完点后图中入度为0的点的个数。第二个问题，可以这么考虑，缩完点后的图中有c1个入度为0的点，有c2个出度为0的点。把入度为0的点和出度为0的点尽量匹配，剩下的就向连通图中连一条边即可，所以第二个问题的答案就是max（c1，c2）。
/** @Author: chenkexing* @Date: ??2018-09-05 11:05:14* @Last Modified by: ??chenkexing* @Last Modified time: 2018-09-07 20:25:39*/#include <algorithm>#include ?<iterator>#include ?<iostream>#include ??<cstring>#include ??<cstdlib>#include ??<iomanip>#include ???<bitset>#include ???<cctype>#include ???<cstdio>#include ???<string>#include ???<vector>#include ????<stack>#include ????<cmath>#include ????<queue>#include ?????<list>#include ??????<map>#include ??????<set>#include ??<cassert>using namespace std;//#pragma GCC optimize(3)//#pragma comment(linker, "/STACK:102400000,102400000") ?//c++#define lson (l , mid , rt << 1)#define rson (mid + 1 , r , rt << 1 | 1)#define debug(x) cerr << #x << " = " << x << "\n";#define pb push_back#define pq priority_queuetypedef long long ll;typedef unsigned long long ull;typedef pair<ll ,ll > pll;typedef pair<int ,int > pii;typedef pair<int,pii> p3;//priority_queue<int> q;//这是一个大根堆q//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q#define fi first#define se second//#define endl ‘\n‘#define OKC ios::sync_with_stdio(false);cin.tie(0)#define FT(A,B,C) for(int A=B;A <= C;++A) ?//用来压行#define REP(i , j , k) ?for(int i = j ; i < ?k ; ++i)//priority_queue<int ,vector<int>, greater<int> >que;const ll mos = 0x7FFFFFFF; ?//2147483647const ll nmos = 0x80000000; ?//-2147483648const int inf = 0x3f3f3f3f; ??????const ll inff = 0x3f3f3f3f3f3f3f3f; //18const int mod = 1e9+7;const double esp = 1e-8;const double PI=acos(-1.0);template<typename T>inline T read(T&x){ ???x=0;int f=0;char ch=getchar(); ???while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); ???while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); ???return x=f?-x:x;}/*-----------------------showtime----------------------*/ ???????????const int maxn = 400; ???????????vector<int>mp[maxn]; ???????????int dfn[maxn],low[maxn],vis[maxn],col[maxn]; ???????????int in[maxn],out[maxn]; ???????????int tot,cnt; ???????????stack<int>S; ???????????void tarjan(int x){ ???????????????low[x] = dfn[x] = ++tot; ???????????????S.push(x);vis[x] = 1; ???????????????for(int i=0; i<mp[x].size(); i++){ ???????????????????int v = mp[x][i]; ???????????????????if(!dfn[v]){ ???????????????????????tarjan(v); ???????????????????????low[x] = min(low[x],low[v]); ???????????????????} ???????????????????else if(vis[v]){ ???????????????????????low[x] = min(low[x], dfn[v]); ???????????????????} ???????????????} ????????????????if(low[x] == dfn[x]){ ???????????????????cnt++; ???????????????????while(true){ ???????????????????????int now = S.top(); ???????????????????????S.pop(); ???????????????????????col[now] = cnt; ???????????????????????vis[now] = 0; ???????????????????????if(now == x)break; ???????????????????} ???????????????} ???????????}int main(){ ???????????int n; ???????????while(~scanf("%d", &n)){ ???????????????for(int i=1; i<=n; i++){ ???????????????????mp[i].clear(); ???????????????????dfn[i] = low[i] = vis[i] = col[i] = 0; ???????????????????in[i] = out[i] = 0; ???????????????????tot = cnt = 0; ???????????????} ???????????????while(!S.empty())S.pop(); ???????????????for(int i=1; i<=n; i++){ ???????????????????int x; ???????????????????while(scanf("%d", &x) && x){ ???????????????????????mp[i].pb(x); ???????????????????} ???????????????} ???????????????for(int i=1; i<=n; i++){ ???????????????????if(dfn[i] == 0){ ???????????????????????tarjan(i); ???????????????????} ???????????????} ???????????????for(int i=1; i<=n; i++){ ???????????????????for(int j=0; j<mp[i].size(); j++){ ???????????????????????int u = i,v = mp[i][j]; ???????????????????????if(col[u] != col[v]){ ???????????????????????????out[col[u]]++; ???????????????????????????in[col[v]]++; ???????????????????????} ???????????????????} ???????????????} ???????????????// debug(cnt); ???????????????????????????????int ans1 = 0,ans2 = 0; ???????????????for(int i=1; i<=cnt; i++){ ???????????????????if(in[i] == 0)ans1++; ???????????????????if(out[i] == 0)ans2++; ???????????????} ???????????????if(cnt==1) ???????????????????printf("1\n0\n"); ???????????????else printf("%d\n%d\n", ans1,max(ans1,ans2)); ???????????} ???????????return 0;}POJ 1236 Network of Schools - 缩点
原文地址：https://www.cnblogs.com/ckxkexing/p/9606950.html