bzoj1148

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1148

很常见的排序贪心题。。。。。。

假设我们得到了一个最优序列，记s[n]=w[1]+w[2]+...+w[n]

对于第n个和第n+1个，剩余容量为：

min(c[n]-s[n-1],c[n+1]-s[n-1]-w[n])

如果交换第n个和第n+1个，剩余容量为：

min(c[n+1]-s[n-1],c[n]-s[n-1]-w[n+1])

因为我们得到的是一个最优序列，所以不交换比交换优，所以：

min(c[n]-s[n-1],c[n+1]-s[n-1]-w[n])  > min(c[n+1]-s[n-1],c[n]-s[n-1]-w[n+1])

容易知道c[n]-s[n-1]>c[n]-s[n-1]-w[n+1]，c[n+1]-s[n-1]-w[n]<c[n+1]-s[n-1]

所以就是：

c[n+1]-s[n-1]-w[n]>c[n]-s[n-1]-w[n+1]
c[n]+w[n]<c[n+1]+w[n+1]

所以就是按ci+wi从小到大排序。

我们先按ci+wi从小到大排序，然后一个一个取，如果不能去，就在已取的中找质量最大的，看替换或会不会更优，这个可以用优先队列。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define re(i,a,b)  for(i=a;i<=b;i++)
#define red(i,a,b) for(i=a;i>=b;i--)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-9;
inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }
inline LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z=='-'){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
        return (neg)?-res:res; 
    }

const int maxN=200000;

int N;
struct Tdata{LL C,W;}data[maxN+100];

inline bool cmp(Tdata a,Tdata b){return a.C+a.W<b.C+b.W;}

struct cmp2{inline bool operator()(Tdata a,Tdata b){return a.W<b.W;}};
priority_queue<Tdata,vector<Tdata>,cmp2> Q;
LL ans;

int main()
  {
      freopen("bzoj1148.in","r",stdin);
      freopen("bzoj1148.out","w",stdout);
      int i;
      N=gint();
      re(i,1,N)data[i].C=LL(gint()),data[i].W=LL(gint());
      sort(data+1,data+N+1,cmp);
      re(i,1,N)
        if(ans<=data[i].C)
          {
              ans+=data[i].W;
              Q.push(data[i]);
          }
        else
          if(data[i].W<Q.top().W && ans-Q.top().W<=data[i].C)
            {
                ans-=Q.top().W;
                Q.pop();
                ans+=data[i].W;
                Q.push(data[i]);
            }
      cout<<Q.size()<<endl;
      cout<<ans<<endl;
      return 0;
  }