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;
}