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G.cpp
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G.cpp
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// https://ncpc14.kattis.com/problems/outing
#include<bits/stdc++.h>
using namespace std;
#define __ ios_base::sync_with_stdio(0); cin.tie(0);
#define endl '\n'
#define foreach(it, x) for (__typeof (x).begin() it = (x).begin(); it != (x).end(); ++it)
#define all(x) x.begin(),x.end()
#define D(x) cout << #x " = " << (x) << endl;
template <class T> string toStr(const T &x)
{ stringstream s; s << x; return s.str(); }
template <class T> int toInt(const T &x)
{ stringstream s; s << x; int r; s >> r; return r; }
int dx[8] = {-1,-1,-1,0,1,1, 1, 0};
int dy[8] = {-1, 0, 1,1,1,0,-1,-1};
int dp[1111][1111][11];
void dfs (int i, vector< vector<int> > &G, vector<int> &visited, int &size) {
if (!visited[i]) {
size ++;
visited[i] = true;
for (int j = 0; j < G[i].size(); ++j) {
int next = G[i][j];
dfs (next, G, visited, size);
}
}
}
int go (int i, int k, bool taken, int sum, vector<int> &toTake) {
if (dp[i][sum][taken] != -1) return dp[i][sum][taken];
if (i >= toTake.size()) return sum;
int a = 0, b = 0;
if (sum + toTake[i] <= k)
a = go (i + 1, k, true, sum + toTake[i], toTake);
b = go (i + 1, k, taken, sum, toTake);
return dp[i][sum][taken] = max(a, b);
}
/** Krustal implementation
* Complexity: O(E log V)
**/
const int mx = 1111;
struct edge {
int start, end, weight;
edge(){} edge(int start, int end, int weight) : start(start), end(end), weight(weight) {}
bool operator < (const edge &that) const {
// Change < by > to find maximum spanning tree
return weight > that.weight;
}
};
/** Begin union find
* Complexity: O(m log n) where m is the number of the operation
* and n are the number of objects, in practice is almost O(m)
**/
int p[mx];
void init_set (int &n) {
for (int i = 0; i <= n; ++i) p[i] = i;
}
int find_set (int x) {
return p[x] == x ? p[x] : p[x] = find_set(p[x]);
}
void join (int x, int y) {
p[find_set(x)] = find_set(y);
}
/** End union find **/
/** e is a vector with all edges of the graph
* return graph of MST
**/
int krustal (vector<edge> &e, int &n, int &k, vector< vector<int> > &G, vector< vector<int> > &G2) {
vector<int> visited (n + 1, 0);
vector<int> visited2 (n + 1, 0);
vector< int > toTake;
int rest = 0;
sort(e.begin(), e.end());
init_set(n);
for (int i = 0 ; i < e.size(); ++i) {
int u = e[i].start;
int v = e[i].end;
int w = e[i].weight;
if (find_set(u) == find_set(v)) {
if (!visited[u]) {
int sizeCycle = 0;
int totalNodes = 0;
dfs (u, G, visited, sizeCycle);
dfs (u, G2, visited2, totalNodes);
if (sizeCycle <= k) {
toTake.push_back(sizeCycle);
rest += totalNodes - sizeCycle;
}
}
}
join(u, v);
}
memset(dp, -1, sizeof dp);
int ans = go (0, k, 0, 0, toTake);
int free = k - ans;
if (rest >= free) ans = k;
else ans += rest;
return ans;
}
int main () {
int n, k, b;
while (cin >> n >> k) {
vector<edge> v(n);
vector< vector<int> > G(n + 1), G2(n + 1);
edge e;
for (int i = 0; i < n; ++i) {
cin >> b;
e.start = i + 1;
e.end = b;
e.weight = 1;
v[i] = e;
G[i + 1].push_back(b);
G2[b].push_back(i + 1);
}
int ans = krustal(v, n, k, G, G2);
cout << ans << endl;
}
}