[NOIP2015 普及组] 推销员

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一道简单的贪心题。

题意

$x$ 轴上有 $n$ 个点（$x_i\in\N^\ast$），每个点有权值 $v_i$，有一小人从原点出发，需要选定 $k$ 个位置进行操作，操作方法是：先走到距离最远的点，操作之，在走回来的路上顺路操作需要操作的点。操作第 $i$ 个点会获得 $v_i$ 的收益，走 $1$ 个单位长度会获得 $1$ 的收益，对每个 $k\in[1,n]$ 分别求出最大收益。

$n\le 10^5$，$x_i\le 10^8$，$v_i\le 1000$。

link: https://www.luogu.com.cn/problem/P2672

source: PJ 2015

做法

容易知道 $k$ 的最优选择（之一）一定是 $k-1$ 的最优选择（之一）的超集。

按 $v_i$ 和 $2x_i+v_i$ 分别维护两个点的 std::set，每次贪心选择即可。

代码

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#include <cstdio>
#include <set>
#include <algorithm>

namespace mirai {

class node {
 public:
  int dist, val;
  friend bool operator<(const node& x, const node& y);
  friend bool operator==(const node& x, const node& y);
} a[100005];

bool operator<(const node& x, const node& y) {
  return x.dist == y.dist ? x.val < y.val : x.dist < y.dist;
}
bool operator==(const node& x, const node& y) {
  return x.dist == y.dist && x.val == y.val;
}

class comp1 {
 public:
  bool operator()(const node& x, const node& y) const {
    return x.val != y.val ? x.val > y.val : x.dist > y.dist;
  }
};
class comp2 {
 public:
  bool operator()(const node& x, const node& y) const {
    return 2 * x.dist + x.val != 2 * y.dist + y.val ? 2 * x.dist + x.val > 2 * y.dist + y.val : !(x < y || x == y);
  }
};
std::multiset<node, comp1> s1;
std::multiset<node, comp2> s2;

int main(int argc, char** argv) {
  int n;
  std::scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    std::scanf("%d", &a[i].dist);
  }
  for (int i = 1; i <= n; ++i) {
    std::scanf("%d", &a[i].val);
  }
  for (int i = 1; i <= n; ++i) {
    s2.insert(a[i]);
  }
  int nowdist = 0, nowidx = 1;
  long long nowans = 0;
  for (int i = 1; i <= n; ++i) {
    // std::printf("s1: ");
    // for (auto i : s1) {
      // std::printf("(%d, %d) ", i.dist, i.val);
    // }
    // std::printf("\ns2: ");
    // for (auto i : s2) {
      // std::printf("(%d, %d) ", i.dist, i.val);
    // }
    // std::printf("\n");
    int maxs2 = -1, maxs1 = -1;
    if (!s2.empty()) {
      auto begin = s2.begin();
      maxs2 = (begin->dist - nowdist) * 2 + begin->val;
    }
    if (!s1.empty()) { maxs1 = s1.begin()->val; }
    if (maxs2 <= maxs1) {
      auto begin = s1.begin();
      nowans += begin->val;
      s1.erase(begin);
    } else {
      auto begin = s2.begin();
      // std::printf("{%d, %d}\n", begin->dist, begin->val);
      nowans += (begin->dist - nowdist) * 2 + begin->val;
      nowdist = begin->dist;
      while (a[nowidx].dist < begin->dist) {
    	  // std::printf("%d %d %s\n", i, nowidx, s2.find(a[nowidx]) == s2.end() ? "YES" : "NO");
        auto find = s2.find(a[nowidx]);
        if (find == s2.end()) {
          nowidx++;
          continue;
        }
        s2.erase(s2.find(a[nowidx]));
        s1.insert(a[nowidx]);
        nowidx++;
      }
      s2.erase(begin);
    }
    std::printf("%lld\n", nowans);
  }
  return 0;
}

} // namespace mirai

int main(int argc, char** argv) {
  return mirai::main(argc, argv);
}