A. Dragons codeforces+贪心
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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 alln dragons that live on this level. Kirito and the dragons havestrength, 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 equalss.
If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strengthxi, 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 byyi.
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.
The first line contains two space-separated integers s andn (1 ≤ s ≤ 104,1 ≤ n ≤ 103). Thenn lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104,0 ≤ yi ≤ 104) — thei-th dragon's strength and the bonus for defeating it.
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.
2 21 99100 0
YES
10 1100 100
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;}
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