題意：給定n個數，Ai的下標為1~n。對于每一個i，Ai與i在同一個樹上，且是與i最遠的點中id最小的點（這個條件變相的說明i與Ai連通）。求森林中樹的個數。

分析：若i與Ai連通，則在同一個樹上，因此連通塊的個數就是樹的個數。并查集即可。

#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 10000 + 10; const int MAXT = 10000 + 10; using namespace std; int fa[MAXN]; int a[MAXN]; set<int> s; int Find(int v){return fa[v] = (fa[v] == v) ? v : Find(fa[v]); } int main(){for(int i = 0; i < MAXN; ++i){fa[i] = i;}int n;scanf("%d", &n);for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);int tx = Find(i);int ty = Find(a[i]);if(tx < ty) fa[ty] = tx;else fa[tx] = ty;}for(int i = 1; i <= n; ++i){if(Find(i) == i) s.insert(i);if(Find(a[i]) == a[i]) s.insert(a[i]);}printf("%d\n", s.size());return 0; }

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轉載于:https://www.cnblogs.com/tyty-Somnuspoppy/p/6294080.html

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