首页 > 代码库 > hdu 3454 Queue-jumpers
hdu 3454 Queue-jumpers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3848 Accepted Submission(s): 1059
Problem Description
Ponyo and Garfield are waiting outside the box-office for their favorite movie. Because queuing is so boring, that they want to play a game to kill the time. The game is called “Queue-jumpers”. Suppose that there are N people numbered from 1 to N stand in a line initially. Each time you should simulate one of the following operations:
1. Top x :Take person x to the front of the queue
2. Query x: calculate the current position of person x
3. Rank x: calculate the current person at position x
Where x is in [1, N].
Ponyo is so clever that she plays the game very well while Garfield has no idea. Garfield is now turning to you for help.
1. Top x :Take person x to the front of the queue
2. Query x: calculate the current position of person x
3. Rank x: calculate the current person at position x
Where x is in [1, N].
Ponyo is so clever that she plays the game very well while Garfield has no idea. Garfield is now turning to you for help.
Input
In the first line there is an integer T, indicates the number of test cases.(T<=50)
In each case, the first line contains two integers N(1<=N<=10^8), Q(1<=Q<=10^5). Then there are Q lines, each line contain an operation as said above.
In each case, the first line contains two integers N(1<=N<=10^8), Q(1<=Q<=10^5). Then there are Q lines, each line contain an operation as said above.
Output
For each test case, output “Case d:“ at first line where d is the case number counted from one, then for each “Query x” operation ,output the current position of person x at a line, for each “Rank x” operation, output the current person at position x at a line.
Sample Input
3
9 5
Top 1
Rank 3
Top 7
Rank 6
Rank 8
6 2
Top 4
Top 5
7 4
Top 5
Top 2
Query 1
Rank 6
Sample Output
Case 1:
3
5
8
Case 2:
Case 3:
3
6
Author
wzc1989
Source
2010 ACM-ICPC Multi-University Training Contest（1）——Host by FZU
还算是比较简单吧，唯一就是注意离散化，将需要的点建一个点，然后将那些不需要的点以区间的形式记录下来。
top删除那个点，然后在将他插入在小的位置。
rank 这个就不说了
query 将这个点放到根节点，然后在输出左边的节点+1.
#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>const int N = 100000 + 11;using namespace std;int t,n,m,tot,li[N],id[N];struct Query{ int ty,num;}quer[N];int root,cnt;struct Splay_tree{ int l,r,cnt,size,f,son[2]; void news(int L,int R) { l = L, r = R, cnt = R - L + 1, size = f = son[0] = son[1] = 0; } void clear() { l = r = cnt = size = f = son[0] = son[1] = 0; }} tree[N<<1];int get(int x){ return x == tree[tree[x].f].son[1];}void updata(int x){ tree[x].size = tree[x].cnt; if(tree[x].son[0]) tree[x].size += tree[tree[x].son[0]].size; if(tree[x].son[1]) tree[x].size += tree[tree[x].son[1]].size;}void rotate(int x){ int fa = tree[x].f; int gfa = tree[fa].f; int ze = get(x); tree[fa].son[ze] = tree[x].son[ze^1]; tree[tree[fa].son[ze]].f = fa; tree[x].son[ze^1] = fa, tree[fa].f = x;tree[x].f = gfa; ze = (fa == tree[gfa].son[1]); if(gfa) tree[gfa].son[ze] = x; updata(fa), updata(x);}void splay(int x){ for(int i; i = tree[x].f;rotate(x)) { if(tree[i].f) if(get(x) == get(i)) rotate(i); } root = x;}void insert(int l,int r,int idx){ //cout<<l<<" "<<r<<" "<<endl; if(root == 0){tree[idx].news(l,r),root = idx; return;} int now = root,fa; while(1) { int ze = l > tree[now].l; fa = now; now = tree[fa].son[ze]; if(!now) { tree[idx].news(l,r);tree[idx].f = fa; tree[fa].son[ze] = idx,updata(idx); updata(fa),splay(idx);return; } }}void Init(){ scanf("%d%d",&n,&m); char ch[10]; for(int i = 1; i <= m; ++i) { scanf("%s%d",ch,&quer[i].num); if(ch[0] == ‘T‘) quer[i].ty = 1; if(ch[0] == ‘R‘) quer[i].ty = 2; if(ch[0] == ‘Q‘) quer[i].ty = 3; li[i] = quer[i].num; } sort(li + 1,li + 1 + m); tot = 1; for(int i = 2; i <= m; ++i) if(li[i] != li[tot]) li[++tot] = li[i]; li[tot+1] = n + 1; cnt = 1; if(li[1] != 1) insert(1,li[1]-1,1); for(int i = 1; i <= tot; ++i) { insert(li[i],li[i],++cnt); id[i] = cnt; if(li[i] != li[i+1] + 1) insert(li[i] + 1, li[i+1] - 1,++cnt); }}int binary_search(int num){ int l = 1,r = tot; while(l <= r) { int mid = l + ((r-l)>>1); if(li[mid] == num) return id[mid]; if(li[mid] < num) l = mid + 1; else r = mid - 1; } return -1;}int pre(int x){ x = tree[x].son[0]; while(tree[x].son[1]) x = tree[x].son[1]; return x;}void del(int x){ splay(x); if(!tree[x].son[0] && !tree[x].son[1]){root = 0, tree[x].clear(); return;} if(!tree[x].son[0] || !tree[x].son[1]) { int ze = !tree[x].son[1]; int root_y = root; root = tree[x].son[ze^1]; tree[root].f = 0, tree[root_y].clear(); return; } int pres = pre(root); splay(pres); tree[pres].son[1] = tree[x].son[1]; tree[tree[pres].son[1]].f = pres; tree[x].clear(),updata(pres);}void insert_min(int l,int idx){ if(root == 0) {tree[idx].news(l,l),root = idx; return;} int now = root; while(1) { int fa = now; now = tree[fa].son[0]; if(!now) { tree[idx].news(l,l); tree[fa].son[0] = idx; tree[idx].f = fa; updata(idx); updata(fa),splay(idx); return; } }}int query(int ranks){ int now = root; while(1) { int sum = 0; if(tree[now].son[0]) sum += tree[tree[now].son[0]].size; if(sum >= ranks) {now = tree[now].son[0]; continue;} if(sum + tree[now].cnt >= ranks) return tree[now].l + ranks - sum - 1; ranks -= (sum + tree[now].cnt),now = tree[now].son[1]; }}void Solve(){ for(int x = 1; x <= m; ++x) { if(quer[x].ty == 1) { int idx = binary_search(quer[x].num); del(idx); insert_min(quer[x].num,idx); } if(quer[x].ty == 3) { int idx = binary_search(quer[x].num); splay(idx); printf("%d\n",tree[tree[idx].son[0]].size + 1); } if(quer[x].ty == 2) printf("%d\n",query(quer[x].num)); }}int main(){ scanf("%d",&t); for(int x = 1; x <= t; ++x) { printf("Case %d:\n",x); cnt = 0; root = 0; memset(tree,0,sizeof(tree)); Init(); Solve(); } return 0;}
hdu 3454 Queue-jumpers
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。