【BZOJ】1620: [Usaco2008 Nov]Time Management 时间管理(贪心)
http://www.lydsy.com/JudgeOnline/problem.php?id=1620
一开始想不通啊。。
其实很简单。。。
每个时间都有个完成时间,那么我们就从最大的 完成时间的开始往前推
每一次更新最早开始时间(min(ans, a[i].y)代表i事件最早的完成时间)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 | #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std; #define rep(i, n) for(int i=0; i<(n); ++i) #define for1(i,a,n) for(int i=(a);i<=(n);++i) #define for2(i,a,n) for(int i=(a);i<(n);++i) #define for3(i,a,n) for(int i=(a);i>=(n);--i) #define for4(i,a,n) for(int i=(a);i>(n);--i) #define CC(i,a) memset(i,a,sizeof(i)) #define read(a) a=getint() #define print(a) printf("%d", a) #define dbg(x) cout << #x << " = " << x << endl #define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; } inline const int getint() { int r=0, k=1; char c= getchar (); for (; c< '0' ||c> '9' ; c= getchar ()) if (c== '-' ) k=-1; for (; c>= '0' &&c<= '9' ; c= getchar ()) r=r*10+c- '0' ; return k*r; } inline const int max( const int &a, const int &b) { return a>b?a:b; } inline const int min( const int &a, const int &b) { return a<b?a:b; } const int N=1005; struct data { int x, y; }a[N]; inline bool cmp( const data &x, const data &y) { return x.y>y.y; } int main() { int n=getint(); for1(i, 1, n) read(a[i].x), read(a[i].y); sort(a+1, a+1+n, cmp); int ans=~0u>>1; for1(i, 1, n) ans=min(ans, a[i].y)-a[i].x; if (ans<0) puts ( "-1" ); else print(ans); return 0; } |
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
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
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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