【解题报告】2017-2018 8th BSUIR Open Programming Contest-C Good subset 线性基+线段树

2017-2018 8th BSUIR Open Programming Contest. Semifinal C题Good subset.

训练上做了三个小时硬肝出来，先写的trie，然后知道铁用线性基，线性基考场现学真刺激hhh

大概题意是n个操作，每个操作往队列队尾加一个数(x属于[0,42])，或者删掉队首的数，每次操作输出能否在已有的数中选出若干个数求得异或和为42

思路：离线操作，时间做区间用线段树维护每个时间点的线性基，从而就避免了线性基的删除操作。但感觉不是正解？？？

线性基的d不能开太大，会卡空间，还有maxn也是。

明天学线性基和高斯消元

线性基板子参考：https://blog.csdn.net/baodream/article/details/83025285

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+7;
struct L_B{int d[8];L_B(){memset(d,0,sizeof(d));}bool insert(int val){for (int i=7;i>=0;i--)if (val&(1<<i)){if (!d[i]){d[i]=val;break;}val^=d[i];}return val>0;}void clear(){memset(d,0,sizeof(d));}
};
L_B merge(const L_B &n1,const L_B &n2)
{L_B ret=n1;for (int i=7;i>=0;i--)if (n2.d[i])ret.insert(n2.d[i]);return ret;
}
struct node{L_B xxj;int left,right;bool tag_judge;L_B tag;
}tree[maxn<<2];
struct lx{int tL,tR;int num;lx(int _tL,int _tR,int _num){tL=_tL;tR=_tR;num=_num;}
};
vector<lx> a;
void build(int root,int left,int right)
{tree[root].left=left,tree[root].right=right;tree[root].tag_judge=false;if(left==right){return;}int mid=(left+right)>>1;build(root<<1,left,mid);build(root<<1|1,mid+1,right);
}
void pushdown(int root)
{if(!tree[root].tag_judge) return;tree[root<<1].tag=merge(tree[root<<1].tag,tree[root].tag),tree[root<<1].tag_judge=true;tree[root<<1].xxj=merge(tree[root<<1].xxj,tree[root].tag);tree[root<<1|1].tag=merge(tree[root<<1|1].tag,tree[root].tag),tree[root<<1|1].tag_judge=true;tree[root<<1|1].xxj=merge(tree[root<<1|1].xxj,tree[root].tag);tree[root].tag.clear(); tree[root].tag_judge=false;
}
void add(int root,int left,int right,int k)
{if(tree[root].left>=left && tree[root].right<=right){tree[root].xxj.insert(k);tree[root].tag.insert(k);tree[root].tag_judge=true;return;}pushdown(root);if(tree[root<<1].right>=left) add(root<<1,left,right,k);if(tree[root<<1|1].left<=right) add(root<<1|1,left,right,k);tree[root].xxj=merge(tree[root<<1].xxj,tree[root<<1|1].xxj);
}
L_B search(int root,int left,int right)
{if(tree[root].left>=left&&tree[root].right<=right) return tree[root].xxj;pushdown(root);L_B sum;sum.clear();if(tree[root<<1].right>=left) sum=merge(sum,search(root<<1,left,right));if(tree[root<<1|1].left<=right) sum=merge(sum,search(root<<1|1,left,right));return sum;
}
void init()
{a.clear();for(int i=0;i<maxn<<2;i++){tree[i].xxj.clear();}
}
int main()
{int n;while(cin>>n){init();queue<int> q; while(!q.empty()) q.pop();queue<int> q2; while(!q2.empty()) q2.pop();char s[5];int num;int start;for(int i=1;i<=n;i++){scanf("%s",s);if(s[0]=='+'){scanf("%d",&num);q.push(num);q2.push(i);}else{num=q.front();q.pop();start=q2.front();q2.pop();a.push_back(lx(start,i-1,num));}}while(!q.empty()){num=q.front(),q.pop();start=q2.front(),q2.pop();a.push_back(lx(start,n,num));}build(1,1,n);for(int i=0;i<a.size();i++){add(1,a[i].tL,a[i].tR,a[i].num);}for(int i=1;i<=n;i++){L_B temp=search(1,i,i);if(temp.insert(42)) printf("No\n");else printf("Yes\n");}    }return 0;
}

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【解题报告】2017-2018 8th BSUIR Open Programming Contest-C Good subset 线性基+线段树

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