1027D. Mouse Hunt：http://codeforces.com/contest/1027/problem/D
题意：
有n个房间，每个房间放置捕鼠器的费用是不同的，已知老鼠在一个房间x，那么他一定会在下一秒到一个特定的房间a【x】。老鼠一开始可能在任意一个房间，问最少需要多少的费用，使得一定能捉到老鼠。
思路：
这道题要在每个环上找一个费用最小的点，放置捕鼠器，进入环的那些点是不用放捕鼠器的。如何在dfs中找到环，并找到最小的点？在dfs中，如果发现下一个点已经走过，就说明遇到环了，那么用u = a[u] 遍历环即可。
#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 = 2e5+9; int a[maxn],n,c[maxn]; int vis[maxn]; int in[maxn]; int dfs(int x,int i){ if(vis[x])return inf; int res = inf; vis[x] = i; if(vis[a[x]] == i){ res = min(res, c[x]); int u = a[x]; while(u!=x){ res = min(res, c[u]); u = a[u]; } } return min(dfs(a[x], i), res); } int main(){ scanf("%d", &n); for(int i=1; i<=n; i++)scanf("%d", &c[i]); for(int i=1; i<=n; i++)scanf("%d", &a[i]); int ans = 0; for(int i=1; i<=n; i++){ int tmp = dfs(i,i); if(tmp <inf) ans += tmp; } printf("%d\n", ans); return 0; }