D. A Game with Traps

time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are playing a computer game, where you lead a party of m soldiers. Each soldier is characterised by his agility ai.

The level you are trying to get through can be represented as a straight line segment from point 0 (where you and your squad is initially located) to point n+1 (where the boss is located).

The level is filled with k traps. Each trap is represented by three numbers li, ri and di. li is the location of the trap, and di is the danger level of the trap: whenever a soldier with agility lower than di steps on a trap (that is, moves to the point li), he gets instantly killed. Fortunately, you can disarm traps: if you move to the point ri, you disarm this trap, and it no longer poses any danger to your soldiers. Traps don’t affect you, only your soldiers.

You have t seconds to complete the level — that is, to bring some soldiers from your squad to the boss. Before the level starts, you choose which soldiers will be coming with you, and which soldiers won’t be. After that, you have to bring all of the chosen soldiers to the boss. To do so, you may perform the following actions:

if your location is x, you may move to x+1 or x−1. This action consumes one second;
if your location is x and the location of your squad is x, you may move to x+1 or to x−1 with your squad in one second. You may not perform this action if it puts some soldier in danger (i. e. the point your squad is moving into contains a non-disarmed trap with di greater than agility of some soldier from the squad). This action consumes one second;
if your location is x and there is a trap i with ri=x, you may disarm this trap. This action is done instantly (it consumes no time).
Note that after each action both your coordinate and the coordinate of your squad should be integers.

You have to choose the maximum number of soldiers such that they all can be brought from the point 0 to the point n+1 (where the boss waits) in no more than t seconds.

Input
The first line contains four integers m, n, k and t (1≤m,n,k,t≤2⋅105, n<t) — the number of soldiers, the number of integer points between the squad and the boss, the number of traps and the maximum number of seconds you may spend to bring the squad to the boss, respectively.

The second line contains m integers a1, a2, …, am (1≤ai≤2⋅105), where ai is the agility of the i-th soldier.

Then k lines follow, containing the descriptions of traps. Each line contains three numbers li, ri and di (1≤li≤ri≤n, 1≤di≤2⋅105) — the location of the trap, the location where the trap can be disarmed, and its danger level, respectively.

Output
Print one integer — the maximum number of soldiers you may choose so that you may bring them all to the boss in no more than t seconds.

Example
inputCopy
5 6 4 14
1 2 3 4 5
1 5 2
1 2 5
2 3 5
3 5 3
outputCopy
3
Note
In the first example you may take soldiers with agility 3, 4 and 5 with you. The course of action is as follows:

go to 2 without your squad;
disarm the trap 2;
go to 3 without your squad;
disartm the trap 3;
go to 0 without your squad;
go to 7 with your squad.
The whole plan can be executed in 13 seconds.

贪心+二分。

对于每一个炸弹，都分为炸弹位置，和拆弹位置。

对于当前前面有炸弹时，我们要考虑拆这个炸弹，拆完之后是继续往后拆，还是直接回去。

这个我们画图就能知道，如果两个炸弹起点和终点位置有交叉，那么就继续拆，否则直接回去。

判断被覆盖就能知道了，然后覆盖我们用前缀和，差分处理一下即可。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=2e5+10;
int m,n,k,tt,a[N],vis[N],L,R,cnt[N];
struct node{int l,r,d;}t[N];
int cmp(int a,int b){return a>b;}
inline int check(int mid){
	int low=a[mid],tot=0; memset(cnt,0,sizeof cnt);
	for(int i=1;i<=k;i++)	if(t[i].d>low){cnt[t[i].l]++,cnt[t[i].r+1]--;}
	for(int i=1;i<=n+1;i++)	cnt[i]+=cnt[i-1];
	for(int i=0;i<=n+1;i++)	tot+=(cnt[i]>=1);
	return tot*2+n+1<=tt;
}
signed main(){
	cin>>m>>n>>k>>tt; L=0,R=m;
	for(int i=1;i<=m;i++)	scanf("%d",&a[i]);
	for(int i=1;i<=k;i++)	scanf("%d %d %d",&t[i].l,&t[i].r,&t[i].d);
	sort(a+1,a+1+m,cmp);
	while(L<R){
		int mid=L+R+1>>1;
		if(check(mid))	L=mid;
		else	R=mid-1; 
	}
	cout<<L<<endl;
	return 0;
}