codeforce452DIV2——E. Segments Removal

题目

Vasya has an array of integers of length n.

Vasya performs the following operations on the array: on each step he finds the longest segment of consecutive equal integers (the leftmost, if there are several such segments) and removes it. For example, if Vasya's array is [13, 13, 7, 7, 7, 2, 2, 2], then after one operation it becomes [13, 13, 2, 2, 2].

Compute the number of operations Vasya should make until the array becomes empty, i.e. Vasya removes all elements from it.

分析

链表+优先队列

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <iostream>
 #include <queue>

 using namespace std;
 const int maxn=+;
 int a[maxn];
 int vis[maxn];

 struct Node{
     int len,id,val,l;
     bool operator <(const Node &rhs)const{
         return len<rhs.len||(len==rhs.len&&l>rhs.l);
     }
 }node[maxn];

 int n;
 int Next[maxn],Last[maxn];
 int main(){
     memset(vis,,sizeof(vis));
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         scanf("%d",&a[i]);
     }
    int cnt=,m=;
     for(int i=;i<=n;i++){
         cnt++;
         if(a[i]!=a[i+]){
             node[++m].len=cnt;
             node[m].id=m;
             node[m].val=a[i];
             node[m].l=i;
             cnt=;
         }
     }
     priority_queue<Node>q;
     //printf("%d",m);

     for(int i=;i<=m;i++){
         Next[i]=i+;
         Last[i]=i-;
         q.push(node[i]);
     }
     Next[m]=;Last[]=;
     int ans=;
     while(!q.empty()){
         while(!q.empty()&&vis[q.top().id])q.pop();
         if(!q.empty()){
         Node x=q.top();q.pop();
         //printf("%d :%d %d\n",x.id,Next[x.id],Last[x.id]);
         //printf("%d %d %d\n",x.id,x.len,x.val);
         ans++;
         Next[Last[x.id]]=Next[x.id];
         Last[Next[x.id]]=Last[x.id];
         if(Last[x.id]&&Next[x.id]&&node[Last[x.id]].val==node[Next[x.id]].val&&!vis[node[Last[x.id]].id]&&!vis[node[Next[x.id]].id]){
                 vis[node[Last[x.id]].id]=;
                 vis[node[Next[x.id]].id]=;

                 int l=node[Next[x.id]].l;
                 node[++m].val=node[Last[x.id]].val;
                 node[m].len=node[Last[x.id]].len+node[Next[x.id]].len;
                 node[m].id=m;
                 node[m].l=l;
                 Next[Last[Last[x.id]]]=m;
                 Last[Next[Next[x.id]]]=m;
                 Next[m]=Next[Next[x.id]];
                 Last[m]=Last[Last[x.id]];
                 q.push(node[m]);
         }
         }
     }
    printf("%d",ans);

 return ;
 }