description

solution

考慮貢獻為正的，顯然花費a最少先做，考慮貢獻為負的，可以將ab調轉過來，那顯然花費最少的先做，也就是b最小的先做。

code

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define ULL unsigned int
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
 

#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=1e5+2;
int t,n;
LL v;
struct rec{
    int p,w;
};
rec a[mn];
bool cmp(rec x,rec y){
    return ((x.w-x.p>=0)&&(y.w-y.p>=0)&&(x.p<y.p))
        ||((x.w-x.p>=0)&&(y.w-y.p<0))
        ||((x.w-x.p<0
 
)&&(y.w-y.p<0)&&(x.w>y.w));
}
int main(){
    freopen("backpack.in","r",stdin);
    freopen("backpack.out","w",stdout);
    scanf("%d",&t);
    fo(cas,1,t){
        scanf("%d%lld",&n,&v);
        fo(i,1,n)scanf("%d%d",&a[i].p,&a[i].w);
        sort(a+1,a+n+1,cmp);
        fo(i,1
 
,n)if(a[i].p<=v)v+=a[i].w-a[i].p;
        else{v=-1;break;}
        if(v!=-1)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}