Codeforces E. Bear and Contribution(枚举维护)

@(E ACMer)

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题意:有n个人都有自己的初始成就值,对每个人可以有发博客(成就+5)和评论(成就+1).两种操作各自的花费时间分别为b和c.要求任意对任意人操纵,让最终至少有k个人有相同的成就值.
分析:

• 首先分两种情况讨论,如果(c∗5<=b),那么b操作就没有存在价值,需要执行A操作.排序之后,肯定是连续的k个(贪心一下容易理解),维护固定k长度的和即可求得.
• 对于两种操作都有会用到的情况就比较复杂了.它甚至都不一定会是排序后的连续k个.但是我们从答案的角度来倒着看就会简化.最终必定会选择一个值作为k个人都相等的值.这个值取余5只有5种情况,我们就枚举这五种情况.
  
  • 对于余数为md的情况,我们仍然要从左到右的维护一个有k个元素的容器,而且每个元素的值是它们先通过评论操作化为余数为md后的值,这样就转化为和第一种情况类似处理了.这个容器气质很符合的显然是优先队列啦….

//注意这里优先队列的仿函数,定义的是后靠优先权,小根堆要用greter<int>, 自定义仿函数的话要用大于符号.priority<int, vector<int>, greater<int> > que;

#include <iostream>#include <cstdio>#include <cstring>#include <set>#include <map>#include <stack>#include <vector>#include <string>#include <queue>#include <cstdlib>#include <cmath>#include <algorithm>#include <ctime>using namespace std;typedef pair<int, int> pii;typedef long long ull;typedef long long ll;typedef vector<int> vi;#define xx first#define yy second#define rep(i, a, n) for (int i = a; i < n; i++)#define sa(n) scanf("%d", &(n))#define vep(c) for(decltype((c).begin()) it = (c).begin(); it != (c).end(); it++)const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e5 + 12;//ll que[maxn * 4];int ct[maxn * 2];//小根堆仿函数class cmp{public:    bool operator()(const ll a, const ll b) {         return a > b;    }};int main(void) {#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    int n, k, b, c;    while (cin >> n >> k >> b >> c) {        rep (i, 0, n) sa(ct[i]), ct[i] += 1e9;        sort(ct, ct + n);        ll ans = 1ll << 60;        if (c * 5 <= b) {             ll sum = 0;             for (int i = 0; i < n; i++) {                 sum += ct[i];                 if (i >= k - 1) {                     ans = min(ans, ((ll)ct[i] * k - sum) * c);                     sum -= ct[i - k + 1];                 }             }        } else {            rep (md, 0, 5) {                ll sum = 0;                //int head = 0, tail = 0;                priority_queue<ll, vector<ll>, cmp> que;                rep (i, 0, n) {                    ll cc = (md + 5 - ct[i] % 5 ) % 5;                    ll bb = (ct[i] + cc) / 5;                    //等效处理之后的值.                    ll cost = bb * b - cc * c;                    sum += cost;                   // while (head == tail || que[tail - 1] > cost) que[tail++] = cost;                    que.push(cost);                    if (i >= k - 1) {                         ans = min(ans, (bb * k * b - sum));                         sum -= que.top();                         que.pop();                    }                }            }        }        cout << ans << endl;    }    return 0;}