CodeForces 230A Dragons(贪心)
2014-07-21 20:19
281 查看
Dragons
time limit per test2 seconds memory limit per test256 megabytes
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.
/*主要是用到了贪心和结构体排序*/
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1005;
int s,n;
typedef struct
{
int x;
int y;
}dragon;
dragon a[maxn];
bool cmp(dragon a,dragon b)
{
return a.x<b.x;/*使结构体按照数据成员x的升序排列*/
}
int main()
{
int i;
while(cin>>s>>n)
{
for(i=0;i<n;i++)
{scanf("%d%d",&a[i].x,&a[i].y);}
sort(a,a+n,cmp);
for(i=0;i<n;i++)
{
if(s>a[i].x){s+=a[i].y;}
else {break;}
}
if(i==n){cout<<"YES"<<endl;}
else {cout<<"NO"<<endl;}
}
return 0;
}
time limit per test2 seconds memory limit per test256 megabytes
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.
/*主要是用到了贪心和结构体排序*/
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1005;
int s,n;
typedef struct
{
int x;
int y;
}dragon;
dragon a[maxn];
bool cmp(dragon a,dragon b)
{
return a.x<b.x;/*使结构体按照数据成员x的升序排列*/
}
int main()
{
int i;
while(cin>>s>>n)
{
for(i=0;i<n;i++)
{scanf("%d%d",&a[i].x,&a[i].y);}
sort(a,a+n,cmp);
for(i=0;i<n;i++)
{
if(s>a[i].x){s+=a[i].y;}
else {break;}
}
if(i==n){cout<<"YES"<<endl;}
else {cout<<"NO"<<endl;}
}
return 0;
}
相关文章推荐
- CodeForces 449A - Jzzhu and Chocolate(贪心)
- codeforces 432E Square Tiling 贪心
- !CodeForces 287B--(贪心、二分)
- CodeForces 443D - Andrey and Problem(贪心)
- Codeforces 486C Palindrome Transformation 贪心
- CodeForces 490E Restoring Increasing Sequence(贪心)
- Codeforces 755B. PolandBall and Game 贪心
- codeforces 675C Money Transfers 贪心
- CodeForces 810B——Summer sell-off ——贪心
- CodeForces 3D Least Cost Bracket Sequence (贪心+优先队列)
- codeforces 540B-贪心
- codeforces 712C C. Memory and De-Evolution(贪心)
- Codeforces 219C Color Stripe 贪心
- CodeForces - 570B 贪心
- codeforces 613B B. Skills(枚举+二分+贪心)
- codeforces 579D D. "Or" Game(前后缀+贪心)
- CodeForces 405A Gravity Flip【贪心】
- Codeforces 158B(贪心问题)
- CodeForces 718A Efim and Strange Grade (贪心)
- Codeforces 46D Parking Lot(贪心模拟)