Building designing UVA - 11039

先取正的和负的绝对值较小者为开头 。然后交替从正负数中取绝对值最小但比上一个大的。

证明：

1.开头选正负数中绝对值较小的：否则能再多放1个。

2.交替选的时候选最小的符合条件的：如果大的符合，换小的仍符合。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
#include <stack>
#include <bitset>
#define mkp make_pair
using namespace std;
const double EPS=1e-8;
typedef long long lon;
const lon SZ=800020,INF=0x7FFFFFFF,mod=1000000007;
int n,res;
struct nd{int val;nd(int a):val(a){}bool operator<(const nd &rbs)const{return abs(val)>abs(rbs.val);}
};priority_queue<nd> pos,neg;void init()
{for(;pos.size();pos.pop());for(;neg.size();neg.pop());cin>>n;for(int i=0;i<n;++i){int tmp;cin>>tmp;if(tmp<0){neg.push(tmp);}else pos.push(tmp);}
}void work()
{res=1;if(pos.size()==0||neg.size()==0)return;int pre;bool prekind=0;int a=neg.top().val,b=pos.top().val;if(abs(a)<abs(b)){prekind=0;pre=abs(a);neg.pop();}else{prekind=1;pre=abs(b);pos.pop();}for(;;){if(prekind==0){for(;pos.size()!=0;){if(abs(pos.top().val)<pre)pos.pop();else break;}if(pos.size()==0)break;else{++res;prekind=1;pre=abs(pos.top().val);pos.pop();}}else{for(;neg.size()!=0;){if(abs(neg.top().val)<pre)neg.pop();else break;}if(neg.size()==0)break;else{++res;prekind=0;pre=abs(neg.top().val);neg.pop();}}}
}int main()
{std::ios::sync_with_stdio(0);//freopen("d:\\1.txt","r",stdin);
    lon casenum;cin>>casenum;//cout<<casenum<<endl;for(lon time=1;time<=casenum;++time){init();work();cout<<res<<endl;}return 0;
}