http://www.elijahqi.win/2018/01/09/codeforces-913d-too-easy-problems/ ‎
D. Too Easy Problems
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are preparing for an exam on scheduling theory. The exam will last for exactly T milliseconds and will consist of n problems. You can either solve problem i in exactly ti milliseconds or ignore it and spend no time. You don’t need time to rest after solving a problem, either.
Unfortunately, your teacher considers some of the problems too easy for you. Thus, he assigned an integer ai to every problem i meaning that the problem i can bring you a point to the final score only in case you have solved no more than ai problems overall (including problem i).
Formally, suppose you solve problems p1, p2, …, pk during the exam. Then, your final score s will be equal to the number of values of j between 1 and k such that k ≤ apj.
You have guessed that the real first problem of the exam is already in front of you. Therefore, you want to choose a set of problems to solve during the exam maximizing your final score in advance. Don’t forget that the exam is limited in time, and you must have enough time to solve all chosen problems. If there exist different sets of problems leading to the maximum final score, any of them will do.
Input
The first line contains two integers n and T (1 ≤ n ≤ 2·105; 1 ≤ T ≤ 109) — the number of problems in the exam and the length of the exam in milliseconds, respectively.
Each of the next n lines contains two integers ai and ti (1 ≤ ai ≤ n; 1 ≤ ti ≤ 104). The problems are numbered from 1 to n.
Output
In the first line, output a single integer s — your maximum possible final score.
In the second line, output a single integer k (0 ≤ k ≤ n) — the number of problems you should solve.
In the third line, output k distinct integers p1, p2, …, pk (1 ≤ pi ≤ n) — the indexes of problems you should solve, in any order.
If there are several optimal sets of problems, you may output any of them.
Examples
Input
5 300
3 100
4 150
4 80
2 90
2 300
Output
2
3
3 1 4
Input
2 100
1 787
2 788
Output
0
0

Input
2 100
2 42
2 58
Output
2
2
1 2
Note
In the first example, you should solve problems 3, 1, and 4. In this case you’ll spend 80 + 100 + 90 = 270 milliseconds, falling within the length of the exam, 300 milliseconds (and even leaving yourself 30 milliseconds to have a rest). Problems 3 and 1 will bring you a point each, while problem 4 won’t. You’ll score two points.
In the second example, the length of the exam is catastrophically not enough to solve even a single problem.
In the third example, you have just enough time to solve both problems in 42 + 58 = 100 milliseconds and hand your solutions to the teacher with a smile.
題意：給定每個 任務的a[]和t[]分別表示 這場考試 整體做了多少題 如果<=a[i]那麼他就可以給我 最後加分+1 如果 >a[i] 就不加分了 那麼 我就每次去貳分一下 我做了多少題 然後 每次去判斷的時候 如果我a[i]< mid那麼就不新增進去 否則新增到佇列裡 然後都選出來之後 貪心的選一下看 如果這麼少的時間我能否湊夠我要的mid個 %%%%icefox 太強啦 qwq我哪會啊 要不是他提醒 我差一點就掉分了
然後最後輸出方案的時候因為我已經知道答案了 再和貳分驗證的時候一樣我去輸出即可

#include<cstdio>
#include<algorithm>
#define N 220000
using namespace std;
int n,T,a[N],t[N],q[N];
struct node{
    int id,t;
}qq[N];
inline bool cmp(node a,node b){return a.t<b.t;} 
inline bool check(int mid){
    int num=0;
    for (int i=1;i<=n;++i) if(a[i]>=mid) q[++num]=t[i];
    sort
 
(q+1,q+num+1);int x=0;if(num<mid) return 0;int time=0;
    for (int i=1;i<=num;++i){
        if(time+q[i]>T) break;
        ++x;time+=q[i];if (x>=mid) return 1;
    }return x>=mid;
}
int main(){
    freopen("cf.in","r",stdin);
    scanf("%d%d",&n,&T);
    for (int i=1;i<=n;++i) scanf("%d%d",&a[i],&t[i]);
    int l=0,r=n;
    while(l<=r){
        int mid=l+r>>1;
        if (check(mid)) l=mid+1;else r=mid-1;
    }printf("%d\n",r);printf("%d\n",r);int num=0;
    for (int i=1;i<=n;++i) if (a[i]>=r) qq[++num].t=t[i],qq[num].id=i;
    sort(qq+1,qq+num+1,cmp);for (int i=1;i<=r;++i) printf("%d ",qq[i].id);
    return 0;
}