CodeForces 230A Dragons

最新推荐文章于 2016-03-01 20:39:30 发布

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最新推荐文章于 2016-03-01 20:39:30 发布

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分类专栏： ACM 文章标签：排序贪心

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本文介绍了一个基于排序和贪心算法的游戏问题解决方案。主人公Kirito需要击败所有龙才能进入游戏的下一关卡。通过将较弱的龙按顺序排列并优先击败，Kirito可以在不损失的情况下增加自己的力量。

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原题传送门：http://codeforces.com/problemset/problem/230/A

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

Note

In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.

In the second sample Kirito's strength is too small to defeat the only dragon and win.

分析：排序+贪心

把那些能被打掉的小怪兽排前面，如果这样都不能顺利通关的话，那肯定不能。

#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 1100
struct Dr{
    int str;
    int bon;
};
bool cmp(Dr a,Dr b){
    return a.str<b.str;
}
int s,n,flag;
Dr d[MAXN];
int main()
{
    while(scanf("%d%d",&s,&n)==2)
    {
        flag = 1;
        for(int i=0;i<n;i++){
            scanf("%d%d",&d[i].str,&d[i].bon);
        }
        sort(d,d+n,cmp);
        for(int i=0;i<n;i++){
            if(s>d[i].str)
            {
                s+=d[i].bon;
            }
            else{
                flag = 0;
                break;
            }
        }
        if(flag){
            printf("YES\n");
        }else{
            printf("NO\n");
        }

    }
    return 0;
}