BZOJ 1620: [Usaco2008 Nov]Time Management 时间管理

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题目

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1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 488  Solved: 296
[Submit][Status]

Description

Ever the maturing businessman, Farmer John realizes that he must manage his time effectively. He has N jobs conveniently numbered 1..N (1 <= N <= 1,000) to accomplish (like milking the cows, cleaning the barn, mending the fences, and so on). To manage his time effectively, he has created a list of the jobs that must be finished. Job i requires a certain amount of time T_i (1 <= T_i <= 1,000) to complete and furthermore must be finished by time S_i (1 <= S_i <= 1,000,000). Farmer John starts his day at time t=0 and can only work on one job at a time until it is finished. Even a maturing businessman likes to sleep late; help Farmer John determine the latest he can start working and still finish all the jobs on time.

N个工作,每个工作其所需时间,及完成的Deadline,问要完成所有工作,最迟要什么时候开始.

Input

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i

Output

* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.

Sample Input

4
3 5
8 14
5 20
1 16

INPUT DETAILS:

Farmer John has 4 jobs to do, which take 3, 8, 5, and 1 units of
time, respectively, and must be completed by time 5, 14, 20, and
16, respectively.

Sample Output

2

OUTPUT DETAILS:

Farmer John must start the first job at time 2. Then he can do
the second, fourth, and third jobs in that order to finish on time.

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题解

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贪心就可以了，每个工作在deadline或者是已知答案两者更早的那个时间完成就可以了。

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代码

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/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL

inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}

const int Maxn=50000;
const int N=100;

int f[Maxn+10];
int p[N+10];
int c[N+10];
int n,h;
int main(){
	scanf("%d%d",&n,&h);
	for (int i=1;i<=n;i++) scanf("%d%d",&p[i],&c[i]);
	memset(f,127,sizeof(f));
	f[0]=0;
	for (int i=1;i<=n;i++){
		for (int v=0;v<=h;v++){
			f[remin(v+p[i],h)]=remin(f[remin(v+p[i],h)],f[v]+c[i]);
		}
	}
	printf("%d\n",f[h]);
	return 0;
}