Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion
like living underground and digging holes.

Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis
has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely
Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.

The first line contains an integer T, indicating the number of test cases. Then follows T cases, each one contains N + 1 lines. The first line contains 2 integers: V, volume of a hole and N, number of equipment respectively. The next
N lines contain N pairs of integers: Ai and Bi.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.

For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".

2

20 3
10 20
3 10
1 7

10 2
1 10
2 11

Yes
No
 
 
 
//本题数据较大，用搜索效率不高，动态规划没能做出来，容易想到贪心；本题最大的问题就是贪心策略的选择；任两种物品thi1(a1,b1),thi2(a2,b2),若先选择thi1,则需x1=max(b1,a1+b2);若先选择thi2,则需x2=max(b2,a2+b1).到底选哪一种需比较x1,x2,选择较小者。若x1<x2,a1+b2<a2+b1,即b2-b1<a2-a1,此时选择thi1;顾贪心策略为先选择差值较大的。
#include<iostream>
#include<algorithm>
using namespace std;
const int MAX=1000+10;
struct node
{
 int a,b;
}per[MAX];
bool cmp(node ta,node tb)
{
 return ta.b-ta.a>tb.b-tb.a;
}
int main()
{
 int t,i,n,v;
 cin>>t;
 while(t--)
 {
  scanf("%d%d",&v,&n);
  for(i=0;i<n;i++)
  {
   scanf("%d%d",&per[i].a,&per[i].b);
  }
  sort(per,per+n,cmp);
  for(i=0;i<n;i++)
  {
   if(v<per[i].b)
    break;
   v-=per[i].a;
  }
  if(i>=n)
   cout<<"Yes"<<endl;
  else 
   cout<<"No"<<endl;
 }
 return 0;
}