Educational Codeforces Round 105 (Rated for Div. 2) C. 1D Sokoban

题解：
我们可以得知，人可以往左边推箱子或者往右边推。
所以我们可以把所有负数放在一起，所有正数放在一起，也就是左边一堆，右边一堆。

然后开始枚举每一个special positions，假设这个点之前的所有箱子都被推到了这个点和这个点之前连续的几个点，然后查询这一段长度覆盖了几个special positions（用二分查找即可），然后再看看这个点之后，是否有箱子就在special positions上，这两个值加起来，取最大的即可。

这里我是用了multiset来水了一下二分，相同操作也可以二分来操作。

代码：

/*Keep on going Never give up*/
//#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#include<iostream>
//#include<string>
//#include<cmath>
//#include<vector>
//#include<algorithm>
//#include<queue>
//#include<string.h>#define int long long
#define endl '\n'using namespace std;
const int maxn=2e5+10;
const int mo=1000;//int a[maxn],b[maxn];int last[maxn];
int sol(multiset<int> a,vector<int>b){int n=b.size();last[n+1]=last[n]=0;for(int i=n-1;i>=0;i--){if(a.count(b[i])) last[i]=last[i+1]+1;else last[i]=last[i+1];}int res=last[0];multiset<int>::iterator it=a.begin();int len=0;for(int i=0;i<n;i++){while(*it<=b[i]&&it!=a.end()){it++;len++;}if(len==0) continue;int val=b[i]-len+1;int pos=lower_bound(b.begin(),b.end(),val)-b.begin();int qian=i-pos+1;int now=qian+last[i];if(a.count(b[i])) now--;res=max(res,now);}return res;
}
signed main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t;cin>>t;while(t--){int n,m;cin>>n>>m;vector<int> y,b;multiset<int> x,a;for(int i=0;i<n;i++){int z;cin>>z;if(z<0) x.insert(-z);else a.insert(z);}for(int i=0;i<m;i++){int z;cin>>z;if(z<0) y.push_back(-z);else b.push_back(z);}reverse(y.begin(),y.end());int ans=0;ans+=sol(x,y);ans+=sol(a,b);cout<<ans<<endl;}}

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