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-;

inline int sgn(DB x){if(abs(x)<EPS)return ;return(x>)?:-;}

const DB Pi=acos(-1.0);

inline int gint()

  {

        int res=;bool neg=;char z;

        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());

        if(z==EOF)return ;

        if(z=='-'){neg=;z=getchar();}

        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());

        return (neg)?-res:res;

    }

inline LL gll()

  {

      LL res=;bool neg=;char z;

        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());

        if(z==EOF)return ;

        if(z=='-'){neg=;z=getchar();}

        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());

        return (neg)?-res:res;

    }

const int maxN=;

int N;

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

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,,N)data[i].C=LL(gint()),data[i].W=LL(gint());

      sort(data+,data+N+,cmp);

      re(i,,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 ;

  }