1620: [Usaco2008 Nov]Time Management 时间管理

1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB

Submit: 506  Solved: 306

[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.

HINT

Source

Silver

 题解：坑爹啊，这次居然CE，吓我一跳——打开一看，代码没复制全TuT。。。突然觉得其实bzoj上面贪心的题才是最令人不敢下手的，看了半天才尝试性的写了个贪心程序，然后碰运气，有时候AC，有时候直接跪。。。书归正传，这个题其实就是先按照deadline时间排个序，然后不断的往前减当前任务消耗的时间，假如出现了减去后小于0的情况，就出-1，还有注意每次减去后到了下一次，然后要和这个新的任务deadline比较小，假如deadline更小的话，则取deadline再减，否则凉拌。。。

 1 type
 2     arr=array[0..2000] of longint;
 3 var
 4    i,j,k,l,m,n:longint;
 5    a,b:arr;
 6 procedure swap(var x,y:longint);
 7           var z:longint;
 8           begin
 9                z:=x;x:=y;y:=z;
10           end;
11 function min(x,y:longint):longint;
12          begin
13               if x<y then min:=x else min:=y;
14          end;
15 procedure sort(l,r:longint);
16           var
17              i,j,x,y:longint;
18           begin
19                i:=l;j:=r;
20                x:=a[(l+r) div 2];
21                repeat
22                      while a[i]<x do inc(i);
23                      while a[j]>x do dec(j);
24                      if I<=j then
25                         begin
26                              swap(a[i],a[j]);
27                              swap(b[i],b[j]);
28                              inc(i);dec(j);
29                         end;
30                until I>j;
31                if l<j then sort(l,j);
32                if i<r then sort(i,r);
33           end;
34 begin
35      readln(n);
36      for i:=1 to n do
37          readln(b[i],a[i]);
38      sort(1,n);
39      l:=a[n];
40      for i:=n downto 1 do
41          begin
42               l:=min(l,a[i]);
43               l:=l-b[i];
44               if l<0 then
45                  begin
46                       writeln(-1);
47                       halt;
48                  end;
49          end;
50      writeln(l);
51 end.
52