BZOJ 1620: [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.

题解:

开始时间有单调性,如果从时间t开始能完成,那么从时间t’<t都能完成。

所以我们可以二分一个t。

在t确定后根据贪心每次做Deadline最靠前的任务，模拟一下。

注意无解输出-1。囧

end.

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
//by zrt
//problem:
using namespace std;
int n;
struct N{
    int t,s;
    friend bool operator < (N a,N b){
        return a.s<b.s;
    }
}a[1005];
int minn=1<<30;
bool judge(int start){
    int t=start;
    for(int i=1;i<=n;i++){
        if(t+a[i].t>a[i].s) return 0;
        else t+=a[i].t;
    }
    return 1;
}
int main(){
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a[i].t,&a[i].s);
        minn=min(minn,a[i].s);
    }
    sort(a+1,a+n+1);
    int l=0,r=minn;
    if(!judge(0)){
        puts("-1");
        goto ed;
    }
    while(r-l>1){
        int m=(l+r)>>1;
        if(judge(m)){
            l=m;
        }else r=m;
    }
    printf("%d\n",l);
    ed:;return 0;
}