[ABC265B] Explore

Problem Statement

Takahashi is exploring a cave in a video game.

The cave consists of $N$ rooms arranged in a row. The rooms are numbered Room $1,2,\ldots,N$ from the entrance.

Takahashi is initially in Room $1$, and the time limit is $T$.

For each $1 \leq i \leq N-1$, he may consume a time of $A_i$ to move from Room $i$ to Room $(i+1)$. There is no other way to move between rooms.
He cannot make a move that makes the time limit $0$ or less.

There are $M$ bonus rooms in the cave. The $i$-th bonus room is Room $X_i$; when he arrives at the room, the time limit increases by $Y_i$.

Can Takahashi reach Room $N$?

Constraints

• $2 \leq N \leq 10^5$
• $0 \leq M \leq N-2$
• $1 \leq T \leq 10^9$
• $1 \leq A_i \leq 10^9$
• $1 < X_1 < \ldots < X_M < N$
• $1 \leq Y_i \leq 10^9$
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$ $M$ $T$
$A_1$ $A_2$ $\ldots$ $A_{N-1}$
$X_1$ $Y_1$
$X_2$ $Y_2$
$\vdots$
$X_M$ $Y_M$

Output

If Takahashi can reach Room $N$, print Yes; otherwise, print No.

---

Sample Input 1

4 1 10
5 7 5
2 10

Sample Output 1

Yes

• Takahashi is initially in Room $1$, and the time limit is $10$.
• He consumes a time of $5$ to move to Room $2$. Now the time limit is $5$. Then, the time limit increases by $10$; it is now $15$.
• He consumes a time of $7$ to move to Room $3$. Now the time limit is $8$.
• He consumes a time of $5$ to move to Room $4$. Now the time limit is $3$.

---

Sample Input 2

4 1 10
10 7 5
2 10

Sample Output 2

No

He cannot move from Room $1$ to Room $2$.

到达一个洞穴，减去路程的价格，加上这里增加的价格，判断有没有时候价格小于等于0.
```cpp
#include
using namespace std;
const int N=1e5+5;
int n,m,x[N],y[N],a[N],p[N];
long long t;
int main()
{
scanf("%d%d%lld",&n,&m,&t);
for(int i=2;i<=n;i++)
scanf("%d",a+i);
for(int i=1;i<=m;i++)
scanf("%d%d",x+i,y+i),p[x[i]]+=y[i];
for(int i=1;i<=n;i++)
{
t-=a[i];
if(t<=0)
{
printf("No");
return 0;
}
t+=p[i];
}
printf("Yes");
return 0;
}
```