Gym 101201I Postman (贪心)

Posted 2020-10-08 dwtfukgv

题意：有个邮递员，要送信，每次最多带 m 封信，有 n 个地方要去送，每个地方有x 封要送，每次都到信全送完了，再回去，对于每个地方，可以送多次直到送够 x 封为止。

析：一个很简单的贪心，就是先送最远的，如果送完最远的还剩下，那么就送次远的，如果不够了，那么就加上回来的距离，重新带够 m 封信，对于左半轴和右半轴都是独立的，两次计算就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
#define pii pair<int, int>
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
  int pos;
  LL val;
  Node(int p = 0, LL v = 0) : pos(p), val(v) { }
  bool operator < (const Node rhs) const{
    return pos > rhs.pos;
  }
};

Node node1[maxn], node2[maxn];

int main(){
  scanf("%d %d", &n, &m);
  int cnt1 = 0, cnt2 = 0;
  for(int i = 0; i < n; ++i){
    int a, b;  scanf("%d %d", &a, &b);
    if (a < 0)  node1[cnt1++] = Node(-a, b);
    else  node2[cnt2++] = Node(a, b);
  }
  sort(node1, node1 + cnt1);
  sort(node2, node2 + cnt2);
  LL ans = 0;
  LL pre_left = 0;
  for(int i = 0; i < cnt1; ++i){
    if (pre_left >= node1[i].val){
      pre_left -= node1[i].val;
      node1[i].val = 0;
      continue;
    }
    node1[i].val -= pre_left;
    LL num = (node1[i].val - 1) / m + 1;
    ans += num * node1[i].pos * 2;
    pre_left = num * m - node1[i].val;
  }
  pre_left = 0;
  for (int i = 0; i < cnt2; ++i){
    if (pre_left >= node2[i].val){
      pre_left -= node2[i].val;
      node2[i].val = 0;
      continue;
    }
    node2[i].val -= pre_left;
    LL num = (node2[i].val - 1) / m + 1;
    ans += num * node2[i].pos * 2;
    pre_left = num * m - node2[i].val;
  }
  printf("%I64d\n", ans);
  return 0;
}