POJ 1236 :http://poj.org/problem?id=1236
参考:https://www.cnblogs.com/TnT2333333/p/6875680.html
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。
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_queue typedef 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; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const 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:https://www.cnblogs.com/ckxkexing/p/9606950.html