HDU3090 Go Home
来源:互联网 发布:关闭windows数字签名 编辑:程序博客网 时间:2024/05/29 06:45
Go Home
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
There comes the holiday, Partychen set foot on the way home. He takes some ECNU coins to hire bodyguards to prevent from being robbed before he went home. But the bodyguard takes one coin for every kilometer. If Partychen walks without bodyguard , he will be robbed one ECNU coin by every robber on every kilometer . Of course , he can choose where to hire bodyguard or where to be robbed as he like.
For example , there are two roads on his way home and he wants to use 8 ECNU coins to hire bodyguard , the first road takes 4 kilometers with 5 robbers ( per kilometer ) and the second takes 5 kilometers with 6 robbers. He could choose the last 3 kilometers on the first road and the whole kilometers on the second road to hire bodyguard to protect him, and leave the first kilometer on the first road to be robbed by 5 robbers, which he will be robbed 5 ECNU coins.
Now , Partychen want to know how many ECNU coins will be robbed at least.
For example , there are two roads on his way home and he wants to use 8 ECNU coins to hire bodyguard , the first road takes 4 kilometers with 5 robbers ( per kilometer ) and the second takes 5 kilometers with 6 robbers. He could choose the last 3 kilometers on the first road and the whole kilometers on the second road to hire bodyguard to protect him, and leave the first kilometer on the first road to be robbed by 5 robbers, which he will be robbed 5 ECNU coins.
Now , Partychen want to know how many ECNU coins will be robbed at least.
Input
It consists of multi-case .
Every case starts with two integers N and M ( 0�N�10,000, 0�M�1,000,000,000 ) which means that there are N roads and M ECNU coins to hire bodyguard.
The followed N lines contains two integers D and P (1<=D<=10,000 , 0<=P<=10 ) , which means the length of every road and the number of robbers in every kilometer on this road.
End with N=0 and M=0 .
Every case starts with two integers N and M ( 0�N�10,000, 0�M�1,000,000,000 ) which means that there are N roads and M ECNU coins to hire bodyguard.
The followed N lines contains two integers D and P (1<=D<=10,000 , 0<=P<=10 ) , which means the length of every road and the number of robbers in every kilometer on this road.
End with N=0 and M=0 .
Output
An integer means the number of ECNU coins to be robbed at least.
Sample Input
2 84 55 6 3 15 105 105 100 0
Sample Output
5 140
题意
回家共有 n (0 <= n <=10,000)段路,有 m (0 <= m <= 1,000,000,000)块钱可以雇佣保镖。每段路有ai(0 <= i < n)千米,每千米有bi(0 <= i < n)个强盗。雇用一个保镖一千米需要花费一块钱,可以保证这一千米内不被强盗抢劫。每千米每个强盗抢一块钱。最少被抢劫的钱数是多少。
分析
以每千米的强盗数降序排序,贪心
AC代码如下
#include <cstdio>#include <algorithm>using namespace std;struct Node{ int d,p;};Node roads[10001];int n,m;int sum;bool cmp(Node n1,Node n2){ return n1.p > n2.p;}int main(){ while(scanf("%d %d", &n, &m), m!=0 || n!=0) { sum = 0; for(int i=0; i<n; i++) { scanf("%d%d", &roads[i].d, &roads[i].p); sum += roads[i].d*roads[i].p; } sort(roads, roads+n, cmp);<span style="white-space:pre"></span>//以强盗数降序sort for(int i=0; i<n; i++)<span style="white-space:pre"></span>//贪心 { if(m > roads[i].d) { m -= roads[i].d; sum -= roads[i].d * roads[i].p; } else { sum -= m * roads[i].p; break; } } printf("%d\n",sum); } return 0;}
0 0
- HDU3090 Go Home
- Go Home.
- Go home!
- go home
- Go Home
- 五一,go home
- Go Home Tonight
- Let me go home :)
- hdu 3090 go home
- 过年GO HOME
- hdu1824Let's go home
- AtCoder Go Home
- I wanna go home
- POJ3767----I Wanna Go Home
- mac home brew install go
- HDU1824 Let's go home
- my home page: skybluemoon.go.nease.com
- go home father teach thoes things
- Mysql导出逗号分隔的csv文件
- 勿忘初心,不断进步
- 关于Unity3d发布后Scene场景的设置(代码控制)
- trace walk DEMO
- Ubantu(Linux)下安装Eclipse并配置 PyDev
- HDU3090 Go Home
- Zend Studio 安装+破解
- 字符和字符串
- poj 3628 Silver Cow Party (spfa)
- 插入排序(InsertSort)
- 最近点对
- 3D游戏《龙心传奇Dragona》全套完整源代码
- Scala隐式转换
- access提示"以独占方式打开或没有权限"的解决方法