A. Dragons codeforces+贪心

最新推荐文章于 2025-01-15 22:05:06 发布

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最新推荐文章于 2025-01-15 22:05:06 发布

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Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.

If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength x_i, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by y_i.

Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.

Input

The first line contains two space-separated integers s and n (1 ≤ s ≤ 10⁴, 1 ≤ n ≤ 10³). Then n lines follow: the i-th line contains space-separated integers x_i and y_i (1 ≤ x_i ≤ 10⁴, 0 ≤ y_i ≤ 10⁴) — the i-th dragon's strength and the bonus for defeating it.

Output

On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.

Sample test(s)

Input

2 2
1 99
100 0

Output

YES

Input

10 1
100 100

Output

NO

解决方案：可将x，y存入结构体，对结构体排序，x优先重小到大，当x相等的时候，y大的优先排序。然后模拟杀怪的过程即可。

代码：#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
int x,y;};
bool cmp(node a,node b){
    if(a.x!=b.x)
        return a.x<b.x;
    else return a.y>b.y;

}
node D[1003];
int s;
int n;
int main(){
    while(~scanf("%d%d",&s,&n)){

        for(int i=0;i<n;i++){
            scanf("%d%d",&D[i].x,&D[i].y);
        }
        sort(D,D+n,cmp);
        int f=1;
        for(int i=0;i<n;i++){
            if(s>D[i].x){
                s+=D[i].y;
            }
            else {f=0;break;}
        }
        if(f)
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }





    return 0;}