【CF1015C】Songs Compression（贪心）

题意：

给定n和m，n组(a[i],b[i])，每一组a[i]可以压缩为b[i]，求最少只需要压缩几个，使得m可以存下所有数据，无解输出-1

思路：按差贪心，排序

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair

 const int N=;

 PII a[N];
 int n,m;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 bool cmp(const PII &a,const PII &b)
 {
     return a.fi-a.se>b.fi-b.se;
 }

 int main()
 {
   //freopen("1.in","r",stdin);
   //freopen("1.out","w",stdout);
   scanf("%d%d",&n,&m);
   ll s=;
   for(int i=;i<=n;i++)
   {
       int x,y;
     scanf("%d%d",&x,&y);
     a[i]=MP(x,y);
     s+=x;
   }
   sort(a+,a+n+,cmp);
   int ans=;
   for(int i=;i<=n;i++)
   {
    if(s>m) {s-=a[i].fi-a[i].se; ans++;}
     else break;
   }
   if(s>m) printf("-1");
    else printf("%d",ans);
   return ;
 }