csu1617]强连通分量

题意：定义域属于一个集合S={0,1,...,n-1}，求S的子集个数，满足以子集的元素为定义域的函数P(x)的值域等于子集本身。

思路：以元素为点，x到P(x)连一条有向边，不难发现，如果有一个有向环，那么环上的元素构成的集合就满足要求。所以问题转化为求有向环的个数，由于有向环之间不可能有交点（同一个点有且仅有一条出边），所以答案就是2^有向环的个数（如果选了有向环上的一点，那么整个有向环必须全部选）。所以只要用tarjan算法统计点数大于等于2的强连通分量个数然后加上自环的，就得到了有向环的个数了。

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <stack>
 16 #include <set>
 17 #include <bitset>
 18 #include <functional>
 19 #include <numeric>
 20 #include <stdexcept>
 21 #include <utility>
 22
 23 using namespace std;
 24
 25 #define mem0(a) memset(a, 0, sizeof(a))
 26 #define mem_1(a) memset(a, -1, sizeof(a))
 27 #define lson l, m, rt << 1
 28 #define rson m + 1, r, rt << 1 | 1
 29 #define define_m int m = (l + r) >> 1
 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 33 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 34 #define all(a) (a).begin(), (a).end()
 35 #define lowbit(x) ((x) & (-(x)))
 36 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 39 #define pchr(a) putchar(a)
 40 #define pstr(a) printf("%s", a)
 41 #define sstr(a) scanf("%s", a)
 42 #define sint(a) scanf("%d", &a)
 43 #define sint2(a, b) scanf("%d%d", &a, &b)
 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 45 #define pint(a) printf("%d\n", a)
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 49 #define mp(a, b) make_pair(a, b)
 50 #define pb(a) push_back(a)
 51
 52 typedef long long LL;
 53 typedef pair<int, int> pii;
 54 typedef vector<int> vi;
 55
 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 58 const int maxn = 1e4 + 7;
 59 const int md = 1e9 + 7;
 60 const int inf = 1e9 + 7;
 61 const LL inf_L = 1e18 + 7;
 62 const double pi = acos(-1.0);
 63 const double eps = 1e-6;
 64
 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 70 int make_id(int x, int y, int n) { return x * n + y; }
 71
 72 struct Graph {
 73     vector<vector<int> > G;
 74     void clear() { G.clear(); }
 75     void resize(int n) { G.resize(n + 2); }
 76     void add(int u, int v) { G[u].push_back(v); }
 77     vector<int> & operator [] (int u) { return G[u]; }
 78 };
 79 Graph G;
 80 int n, m;
 81 int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;
 82 stack<int> S;
 83 int cnt[maxn], a[maxn];
 84
 85 void dfs(int u) {
 86     pre[u] = lowlink[u] = ++ dfs_clock;
 87     S.push(u);
 88     rep_up0(i, G[u].size()) {
 89         int v = G[u][i];
 90         if (!pre[v]) {
 91             dfs(v);
 92             min_update(lowlink[u], lowlink[v]);
 93         }
 94         else if (!sccno[v]) {
 95             min_update(lowlink[u], pre[v]);
 96         }
 97     }
 98     if (lowlink[u] == pre[u]) {
 99         scc_cnt ++;
100         for(;; ) {
101             int x = S.top(); S.pop();
102             sccno[x] = scc_cnt;
103             if (x == u) break;
104         }
105     }
106 }
107 int find_scc(int n) {
108     dfs_clock = scc_cnt = 0;
109     mem0(sccno);
110     mem0(pre);
111     rep_up0(i, n) {
112         if (!pre[i]) dfs(i);
113     }
114     mem0(cnt);
115     int c = 0;
116     rep_up0(i, n) {
117         cnt[sccno[i]] ++;
118     }
119     rep_up1(i, scc_cnt) {
120         if (cnt[i] >= 2) c ++;
121     }
122     rep_up0(i, n) if (G[i].size() == 0) c ++;
123     int ans = 1;
124     rep_up0(i, c) {
125         ans = (ans << 1) % md;
126     }
127     return ans;
128 }
129
130 int P(int x) {
131     int ans = 0;
132     rep_up0(i, m + 1) {
133         ans = (ans * x + a[m - i]) % n;
134     }
135     return ans;
136 }
137
138 int main() {
139     //freopen("in.txt", "r", stdin);
140     int T;
141     cin >> T;
142     while (T --) {
143         cin >> n >> m;
144         G.clear();
145         G.resize(n);
146         rep_up0(i, m + 1) sint(a[i]);
147         rep_up0(i, n) {
148             int x = P(i);
149             if (x != i) G.add(i, x);
150         }
151         cout << find_scc(n) << endl;
152     }
153     return 0;
154 }