Bzoj 1620: [usaco Nov] Time Management

Description

Ever the maturing businessman, Farmer John realizes that he must manage his time was tively. 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 when tively, 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 jobs, the time required for each job, and the deadline to be completed, ask when to start. 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.

Question:

The start time is monotonically updated. If it can be completed from time t, it can be completed from time t' <t.

So we can divide it into two TBS.

After T is determined, perform the top deadline Task Based on greed and simulate it.

Note that there is no solution output-1. Bytes

End.

Code:

#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;}

Bzoj 1620: [usaco Nov] Time Management