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]
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; }