2016 Winter Training Day #1_A题_codefcrces 230A(贪心)

A. Dragons

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 21 99100 0

output

YES

input

10 1100 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 to2 + 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.

题意：给出主角的初始能力为s, 还有n个怪兽，每个怪兽的能力为xi, 每战胜一个怪兽，主角的能力就可以提升yi。问最后主角能否消灭所有怪兽。

思路：简单贪心，将n个怪兽按能力升序排序，然后循环一遍，如果在某次判断中s < xi，就不能完成，否则s累加yi。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;const int INF = 0x7fffffff;struct Node{    int x, y;    Node():x(0), y(0){};};Node a[1200];bool cmp(Node a, Node b){    return a.x < b.x;}int main(){    int s, n;    cin >> s >> n;    for(int i = 0; i < n; ++i)    {        cin >> a[i].x >> a[i].y;    }    sort(a, a+n, cmp);    int ok = 1;    for(int i = 0; i < n; ++i)    {        if(s > a[i].x)        {            s += a[i].y;        }        else        {            ok = 0;            break;        }    }    if(ok)        cout << "YES" << endl;    else        cout << "NO" << endl;    return 0;}