Dragons

Dragons
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 xi, 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 yi.

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 xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 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.

題意就是一個人殺龍，這個人初始力量為s，有n條龍，可以按任何順序殺龍，殺死一條龍有力量獎勵y，龍的力量為x，當這個人的力量s大於龍的力量x時才能殺死這條龍，然後他的力量增加這條龍的獎勵y，即s+=y
所以直接貪心就行了。

AC代碼：

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

struct Node
{
    int x,y;
}g[1010];

bool cmp(Node a, Node b)
{
    return a.x < b.x;
}

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

    return 0;
}