http://www.lydsy.com/JudgeOnline/problem.php?id=2020

和背包差不多

同样滚动数组

f[j]表示当前位置j份食物的最小价值

f[j]=min(f[j-l]+l*c) 1<=l<=f

而且在每一步走的时候

f[j]+=j

然后就行了。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=105, K=105, E=400;
int n, k, e;
int f[K];
struct dat { int x, f, c; }a[N]; int main() {
read(k); read(e); read(n);
for1(i, 1, n) read(a[i].x), read(a[i].f), read(a[i].c);
CC(f, 0x3f); f[0]=0;
for1(p, 0, e) {
for1(j, 1, k) f[j]+=j;
for1(i, 1, n) if(a[i].x==p) {
for3(j, k, 1) {
for1(l, 1, a[i].f) if(j-l>=0)
f[j]=min(f[j], f[j-l]+l*a[i].c);
}
}
}
print(f[k]);
return 0;
}

噗,看了题解后是贪心。。自己想想也是显然啊。。

如果在这个商店买了后,一直要运到终点啊,所以将物品全部拆开后算距离然后贪心取就行了。。


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 D*K 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 1*1 = 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

Input

第1行:K,E,N 第2~N+1行:X_i,F_i,C_i.

Output

Sample Input

2 5 3
3 1 2
4 1 2
1 1 1

Sample Output

7

HINT

Source

【BZOJ】2020: [Usaco2010 Jan]Buying Feed, II (dp)的更多相关文章

  1. 【BZOJ】2101: [Usaco2010 Dec]Treasure Chest 藏宝箱(dp)

    http://www.lydsy.com/JudgeOnline/problem.php?id=2101 这个dp真是神思想orz 设状态f[i, j]表示i-j先手所拿最大值,注意,是先手 所以转移 ...

  2. 2020: [Usaco2010 Jan]Buying Feed, II

    2020: [Usaco2010 Jan]Buying Feed, II Time Limit: 3 Sec  Memory Limit: 64 MBSubmit: 220  Solved: 162[ ...

  3. BZOJ 2020 [Usaco2010 Jan]Buying Feed,II:贪心【定义价值】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=2020 题意: 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 ...

  4. 【BZOJ】3433: [Usaco2014 Jan]Recording the Moolympics (贪心)

    http://www.lydsy.com/JudgeOnline/problem.php?id=3433 想了好久啊....... 想不出dp啊......sad 后来看到一英文题解......... ...

  5. 【BZOJ】2102: [Usaco2010 Dec]The Trough Game(暴力)

    http://www.lydsy.com/JudgeOnline/problem.php?id=2102 直接枚举所有情况......然后判断是否可行.. #include <cstdio> ...

  6. 【BZOJ】1649: [Usaco2006 Dec]Cow Roller Coaster(dp)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1649 又是题解... 设f[i][j]表示费用i长度j得到的最大乐趣 f[i][end[a]]=ma ...

  7. 【BZOJ】1828: [Usaco2010 Mar]balloc 农场分配(经典贪心)

    [算法]贪心+线段树 [题意]给定n个数字ci,m个区间[a,b](1<=a,b<=10^5),每个位置最多被ci个区间覆盖,求最多选择多少区间. 附加退化问题:全部ci=1,即求最多的不 ...

  8. 【BZOJ】2021: [Usaco2010 Jan]Cheese Towers(dp)

    http://www.lydsy.com/JudgeOnline/problem.php?id=2021 噗,自己太弱想不到. 原来是2次背包. 由于只要有一个大于k的高度的,而且这个必须放在最顶,那 ...

  9. BZOJ2020: [Usaco2010 Jan]Buying Feed II

    [传送门:BZOJ2020] 简要题意: 约翰开车回家,遇到了双十一节,那么就顺路买点饲料吧.回家的路程一共有E 公里,这一路上会经过N 家商店,第i 家店里有Fi 吨饲料,售价为每吨Ci 元.约翰打 ...

随机推荐

  1. 浅析SQL Server中的执行计划缓存(上)

    简介 我们平时所写的SQL语句本质只是获取数据的逻辑,而不是获取数据的物理路径.当我们写的SQL语句传到SQL Server的时候,查询分析器会将语句依次进行解析(Parse).绑定(Bind).查询 ...

  2. JMeter 十四:最佳实践

    参考:http://jmeter.apache.org/usermanual/best-practices.html 1. 总是使用最新版本的JMeter 2. 使用合适数目的Thread Threa ...

  3. google 访问技术

    空闲时间提供一些关于google访问的技术分享及技术支持. 不卖产品,请不要询问. 探讨技术请加群.

  4. javaScript Windows相关

    javaScript 关于Windows 1 Windows 对象 <1>全部浏览器都支持 window 对象.它表示浏览器窗体. <2>全部 JavaScript 全局对象. ...

  5. Docker exec与Docker attach

    转载博客地址:http://blog.csdn.net/halcyonbaby 新浪微博:@寻觅神迹 内容系本人学习.研究和总结,如有雷同,实属荣幸! ================== Docke ...

  6. sql 语句中 id&lt ;SELECT * FROM t_blog WHERE id&lt;#{id} ORDER BY id DESC LIMIT 1

  7. 插入排序(PHP,C)

    PHP<?php /* ** 功能:插入算法 ** 描述: ** 作者:yuwensong */ function insertSorting($arr,$n){ for($i = 1; $i& ...

  8. XML - 十分钟了解XML结构以及DOM和SAX解析方式

    引言 NOKIA 有句著名的广告语:"科技以人为本".不论什么技术都是为了满足人的生产生活须要而产生的.详细到小小的一个手机.里面蕴含的技术也是浩如烟海.是几千年来人类科技的结晶, ...

  9. MEF教程

    http://www.cnblogs.com/content/archive/2013/05/31/3111156.html

  10. Android开发-状态栏着色原理和API版本号兼容处理

    介绍 先上实际效果图,有三个版本号请注意区分API版本号 API>=20 API=19 API<19 watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZX ...