BZOJ 1620: [Usaco2008 Nov]Time Management 时间管理( 二分答案 )

二分一下答案就好了...

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cctype>

#include<iostream>

 

#define rep( i , n ) for( int i = 0 ;  i < n ; ++i )

#define clr( x , c ) memset( x , c , sizeof( x ) )

 

using namespace std;

 

const int maxn = 1000 + 5;

 

int n;

 

struct data {

int t , s;

void Read() {

scanf( "%d%d" , &t , &s );

}

bool operator < ( const data &rhs ) const {

return s < rhs.s;

}

};

 

data A[ maxn ];

 

bool jud( int cur ) {

rep( i , n ) {

data &x = A[ i ];

if( ( cur += x.t ) > x.s ) return false;

}

return true;

}

 

int main() {

// freopen( "test.in" , "r" , stdin );

cin >> n;

int L = 0 , R = 0x7fffffff;

rep( i , n ) {

   A[ i ].Read();

   

   R = min( A[ i ].s , R );

   

}

sort( A , A + n );

int ans = -1;

while( L <= R ) {

int mid = ( L + R ) >> 1;

if( jud( mid ) ) 

   ans = mid , L = mid + 1;

   

else 

 

   R = mid - 1;

}

cout << ans << "\n";

return 0;

}

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1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 539  Solved: 325
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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