C. Songs Compression
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Ivan has nn songs on his phone. The size of the ii-th song is aiai bytes. Ivan also has a flash drive which can hold at most mm bytes in total. Initially, his flash drive is empty.
Ivan wants to copy all nn songs to the flash drive. He can compress the songs. If he compresses the ii-th song, the size of the ii-th song reduces from aiai to bibi bytes (bi<aibi<ai).
Ivan can compress any subset of the songs (possibly empty) and copy all the songs to his flash drive if the sum of their sizes is at most mm. He can compress any subset of the songs (not necessarily contiguous).
Ivan wants to find the minimum number of songs he needs to compress in such a way that all his songs fit on the drive (i.e. the sum of their sizes is less than or equal to mm).
If it is impossible to copy all the songs (even if Ivan compresses all the songs), print "-1". Otherwise print the minimum number of songs Ivan needs to compress.
Input
The first line of the input contains two integers nn and mm (1≤n≤105,1≤m≤1091≤n≤105,1≤m≤109) — the number of the songs on Ivan's phone and the capacity of Ivan's flash drive.
The next nn lines contain two integers each: the ii-th line contains two integers aiai and bibi (1≤ai,bi≤1091≤ai,bi≤109, ai>biai>bi) — the initial size of the ii-th song and the size of the ii-th song after compression.
Output
If it is impossible to compress a subset of the songs in such a way that all songs fit on the flash drive, print "-1". Otherwise print the minimum number of the songs to compress.
Examples
input
Copy
4 21
10 8
7 4
3 1
5 4
output
Copy
2
input
Copy
4 16
10 8
7 4
3 1
5 4
output
Copy
-1
Note
In the first example Ivan can compress the first and the third songs so after these moves the sum of sizes will be equal to 8+7+1+5=21≤218+7+1+5=21≤21. Also Ivan can compress the first and the second songs, then the sum of sizes will be equal 8+4+3+5=20≤218+4+3+5=20≤21. Note that compressing any single song is not sufficient to copy all the songs on the flash drive (for example, after compressing the second song the sum of sizes will be equal to 10+4+3+5=22>2110+4+3+5=22>21).
In the second example even if Ivan compresses all the songs the sum of sizes will be equal 8+4+1+4=17>168+4+1+4=17>16.
算法分析：
贪心选择+sum1的提前记录
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=5010;
const int MAXM=100010;
using namespace std;
struct node
{
int a,b,c;
}x[100005];
bool cmp(const node & w,const node &y)
{
return w.c>y.c;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
long long sum=0,sum1=0;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&x[i].a,&x[i].b);
sum+=x[i].b;
x[i].c=x[i].a-x[i].b;
sum1+=x[i].a;
}
if(sum>m)
{
cout<<-1<<endl;
continue;
}
if(sum1<=m)
{
cout<<0<<endl;
continue;
}
sort(x+1,x+n+1,cmp);
int flag=0;
int i;
for( i=1;i<=n;i++)
{
sum1-=x[i].c;
if(sum1<=m) break;
}
printf("%d\n",i);
}
return 0;
}
