[BZOJ2020]&&[Usaco2010 Jan]Buying Feed, II

描述 Description

(buying.pas/buying.in/buying.out 128M 1S) Farmer John needs to travel to town to pick up K (1 <= K <= 100) pounds of feed. Driving D miles with K pounds of feed in his truck costs DK cents. The county feed lot has N (1 <= N <= 100) stores (conveniently numbered 1..N) that sell feed. Each store is located on a segment of the X axis whose length is E (1 <= E <= 350). Store i is at location X_i (0 < X_i < E) on the number line and can sell FJ as much as F_i (1 <= F_i <= 100) pounds of feed at a cost of C_i (1 <= C_i <= 1,000,000) cents per pound. Amazingly, a given point on the X axis might have more than one store. FJ starts at location 0 on this number line and can drive only in the positive direction, ultimately arriving at location E, with at least K pounds of feed. He can stop at any of the feed stores along the way and buy any amount of feed up to the the store’s limit. What is the minimum amount FJ has to pay to buy and transport the K pounds of feed? FJ knows there is a solution. Consider a sample where FJ needs two pounds of feed from three stores (locations: 1, 3, and 4) on a number line whose range is 0..5: 0 1 2 3 4 5 +—|—+—|—|—+ 1 1 1 Available pounds of feed 1 2 2 Cents per pound It is best for FJ to buy one pound of feed from both the second and third stores. He must pay two cents to buy each pound of feed for a total cost of 4. When FJ travels from 3 to 4 he is moving 1 unit of length and he has 1 pound of feed so he must pay 11 = 1 cents. When FJ travels from 4 to 5 he is moving one unit and he has 2 pounds of feed so he must pay 1*2 = 2 cents. The total cost is 4+1+2 = 7 cents.

FJ 开车去买 K 份食物,如果他的车上有 X 份食物。每走一里就花费 X 元。 FJ 的城市是一条线,总共 E 里路,有 E+1 个地方,标号 0~E。 FJ 从 0 开始走,到 E 结束(不能往回走),要买 K 份食物。 城里有 N 个商店,每个商店的位置是 X_i(一个点上可能有多个商店),有 F_i 份食物,每份 C_i 元。 问到达 E 并买 K 份食物的最小花费

输入格式 InputFormat

第 1 行:K,E,N 第 2~N+1 行:X_i,F_i,C_i.

输出格式 OutputFormat

到达 E 并买 K 份食物的最小花费.

样例输入 SampleInput

10 5 5
4 46 1
2 25 1
1 39 1
4 48 3
2 44 4

样例输出 SampleOutput


BZOJ 2020

代码 Code

将 xi 点的食物费用加上 e-xi 后贪心即可。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int i,j,t,n,m,l,r,k,z,y,x;
inline void read(int &x)
    x=0;char ch=getchar();
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
struct food
    int f,c;
    bool operator < (const food &temp) const
        return (c==temp.c)?(f<temp.f):(c<temp.c);
int e,ans;
int main()
    for (i=1;i<=n;i++) read(x),read(a[i].f),read(a[i].c),a[i].c+=e-x;
    while (k)
    return 0;