Codeforces 755F 规模较大的装箱dp

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http://codeforces.com/problemset/problem/755/F

max只要仔细一点就好了，关键是求min

1.因为送礼数组是1-n的排列，所以首先把所有送礼的圈找出来。

2.简单贪心后，min只有可能是K或者K+1，到底是哪一个就看有没有某些圈的size和刚好是K

3.然后好像就变成了一个(1e6*1e6)的装箱dp。就算bitset优化复杂度也在1e10以上。

4.题解就是一个套路: size大的圈bitset优化做，size小的中一样的放一起做多重背包(当然这个多重背包要是O(k)的)……

5.玄学……感受一下学个思想吧

6.还有分块标准也不总是sqrt，居然选什么做临界要看复杂度的表达式……

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <set>#include <map>#include <queue>#include <algorithm>#include <stack>#include <cctype>#include <cmath>#include <vector>#include <sstream>#include <bitset>#include <deque>#include <iomanip>using namespace std;#define pr(x) cout << #x << " = " << x << endl;#define bug cout << "bugbug" << endl;#define ppr(x, y) printf("(%d, %d)\n", x, y);#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define SQR(a) ((a)*(a))#define PCUT puts("\n---------------")typedef long long ll;typedef double DBL;typedef pair<int, int> P;typedef unsigned int uint;const int MOD = 1e9 + 7;const int inf = 0x3f3f3f3f;const ll INF = 0x3f3f3f3f3f3f3f3f;const int maxn = 1e6 + 4;const int maxm = 1e2 + 4;const double pi = acos(-1.0);int n, K;int a[maxn];bool vis[maxn]; vector<int> sz;int cnt[maxm], lst[maxm];bitset<maxn> dp;int main(){//必须编译过才能交int ik, i, j, k, kase;scanf("%d%d", &n, &K);for (i = 1; i <= n; ++i) scanf("%d", a+i);for (i = 1; i <= n; ++i)if (!vis[i]){int cnt = 0;int beg = i;do{cnt++;beg = a[beg];vis[beg] = true;}while(beg != i);sz.push_back(cnt);}int maxv = 0;if (K <= sz.size()) maxv = 2 * K;else if (K >= n - sz.size()) maxv = n;else{int tot = 0, res = 0;for (i = 0; i < sz.size(); ++i) tot += sz[i] - sz[i] % 2, res += sz[i]&1; if (tot >= K*2) maxv = K * 2;else if (res >= (K-tot/2)) maxv = tot + K - tot/2;else maxv = n;}dp.reset();dp[0] = 1;for (i = 0; i < sz.size(); ++i)if (sz[i] >= 100)dp |= dp << sz[i];else cnt[sz[i]]++;for (i = 2; i < 100; ++i)if (cnt[i]){memset(lst, 0x3f, sizeof lst);for (j = 0; j <= K; ++j){int index = j % i;if (dp[j]) lst[index] = j;else if (lst[index] != inf && (j - lst[index]) / i <= cnt[i]) dp[j] = true;}}int minv = dp[K] ? K : K + 1;printf("%d %d\n", minv, maxv);return 0;}

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