bzoj1620 [Usaco2008 Nov]Time Management 时间管理

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.

水题一道……

首先按deadline降序排一遍，保存一个当前的最大开始时间，那么当加入一个工作，要么保存答案的比这个工作的deadline大，要么后者大。无论如何一定要保证做完这个工作之后的时间比两者都大，这样才满足条件。所以两者取小的就行了。不会的自己再yy吧……

#include<cstdio>

#include<algorithm>

using namespace std;

struct work{

	int s,t;

}a[1010];

int n,ans=10000000;

inline int read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

inline int min(int a,int b)

{return a<b?a:b;}

inline bool cmp(const work &a,const work &b)

{return a.s>b.s;}

int main()

{

	n=read();

	for (int i=1;i<=n;i++)

	{

		a[i].t=read();

		a[i].s=read();

	}

	sort(a+1,a+n+1,cmp);

	for (int i=1;i<=n;i++)

	  ans=min(ans,a[i].s)-a[i].t;

	if (ans<0) ans=-1;

	printf("%d",ans);

}

巴特西