Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) A 题目忘了
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题目大意:
反正就是字符串覆盖,但是已经被覆盖了就不会再被覆盖了,问最终的字符串。
非常显然可以用链表/并查集/线段树做。这是比较显然的做法。这里我用优先队列做了一发,效果还不错。
首先需要了解堆的删除操作和离线区间加。那么我就用离线区间加的思想,开一个vector,在开始位置加入插入节点,在结束位置加入删除节点。由于修改是有顺序的,那么可以通过调用优先队列的队头来用最早的字符串覆盖。做法非常显然,就不详细描述了。
附代码:
#include<bits/stdc++.h>#define N 2000000using namespace std;typedef pair<int,int> pii;struct Q{priority_queue<pii,vector<pii>,greater<pii> > q1,q2;inline void check(){while(!q2.empty()&&q1.top().first==q2.top().first) q1.pop(),q2.pop();}inline bool empty(){check();return q1.empty();}inline void push(pii a){check();q1.push(a);}inline void erase(pii a){check();q2.push(a);}inline pii front(){check();return q1.top();}}q;inline void read(int &a){int f=a=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=1;ch=getchar();}while(ch<='9'&&ch>='0') a=a*10+ch-'0',ch=getchar();if(f) a=-a;}struct Node{pii a;bool f;};int n;string s[N+100];char ans[N+100];vector<Node> qz[N+100];int l=1;int main(){scanf("%d",&n);for(int i=0,k;i<n;i++){cin>>s[i];read(k);int t=s[i].size();for(int j=0,a;j<k;j++){read(a);l=max(l,a+t);qz[a].push_back((Node){pii(i,a),1});qz[a+t].push_back((Node){pii(i,a),0});}}for(int i=1;i<l;i++){int t=qz[i].size();for(int j=0;j<t;j++){if(qz[i][j].f) q.push(qz[i][j].a);else q.erase(qz[i][j].a);}if(!q.empty()) ans[i]=s[q.front().first][i-q.front().second];else ans[i]='a';}printf("%s\n",ans+1);return 0;}
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