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
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 #include <iostream> 5 #include <queue> 6 7 using namespace std; 8 const int maxn=200000+10; 9 int a[maxn]; 10 int vis[maxn]; 11 12 struct Node{ 13 int len,id,val,l; 14 bool operator <(const Node &rhs)const{ 15 return len<rhs.len||(len==rhs.len&&l>rhs.l); 16 } 17 }node[maxn]; 18 19 int n; 20 int Next[maxn],Last[maxn]; 21 int main(){ 22 memset(vis,0,sizeof(vis)); 23 scanf("%d",&n); 24 for(int i=1;i<=n;i++){ 25 scanf("%d",&a[i]); 26 } 27 int cnt=0,m=0; 28 for(int i=1;i<=n;i++){ 29 cnt++; 30 if(a[i]!=a[i+1]){ 31 node[++m].len=cnt; 32 node[m].id=m; 33 node[m].val=a[i]; 34 node[m].l=i; 35 cnt=0; 36 } 37 } 38 priority_queue<Node>q; 39 //printf("%d",m); 40 41 for(int i=1;i<=m;i++){ 42 Next[i]=i+1; 43 Last[i]=i-1; 44 q.push(node[i]); 45 } 46 Next[m]=0;Last[1]=0; 47 int ans=0; 48 while(!q.empty()){ 49 while(!q.empty()&&vis[q.top().id])q.pop(); 50 if(!q.empty()){ 51 Node x=q.top();q.pop(); 52 //printf("%d :%d %d ",x.id,Next[x.id],Last[x.id]); 53 //printf("%d %d %d ",x.id,x.len,x.val); 54 ans++; 55 Next[Last[x.id]]=Next[x.id]; 56 Last[Next[x.id]]=Last[x.id]; 57 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]){ 58 vis[node[Last[x.id]].id]=1; 59 vis[node[Next[x.id]].id]=1; 60 61 int l=node[Next[x.id]].l; 62 node[++m].val=node[Last[x.id]].val; 63 node[m].len=node[Last[x.id]].len+node[Next[x.id]].len; 64 node[m].id=m; 65 node[m].l=l; 66 Next[Last[Last[x.id]]]=m; 67 Last[Next[Next[x.id]]]=m; 68 Next[m]=Next[Next[x.id]]; 69 Last[m]=Last[Last[x.id]]; 70 q.push(node[m]); 71 } 72 } 73 } 74 printf("%d",ans); 75 76 return 0; 77 }