【BZOJ】1620: [Usaco2008 Nov]Time Management 时间管理(贪心)




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)

#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;






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.



* Line 1: A single integer: N

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


* 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


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



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.



posted @ 2014-09-05 05:36  iwtwiioi  阅读(272)  评论(0编辑  收藏  举报