http://poj.org/problem?id=2828
一开始敲了个splay，直接模拟。
tle了。。
常数太大。。
好吧，说是用线段树。。
而且思想很拽。。
（貌似很久以前写过貌似的，，）
我们线段树维护的区间不再是人了。。
而是这个区间剩余的的座位。。
比如我现在要坐第一张，但是人已经坐了，即这个区间已经没有位置了。。那就要往后坐。
所以我们逆序添加，，因为后来人插队前边人管不着。。。
所以后来人一定是先定座位的。。
每一次维护这个座位区间。。
如果左边这个区间座位比我要坐的座位号要多（即我要坐的这个座位前边有空位（包括这个））
那么我就往前坐，否则向后坐。。
splay：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=200005;
int n;
struct node* null;
struct node {
int key, s;
node *ch[2], *fa;
node(const int &_k=0, const int &_s=1) : key(_k), s(_s) { ch[0]=ch[1]=fa=null; }
inline void setc(node* c, const bool d) { ch[d]=c; c->fa=this; }
inline int d() { return fa->ch[1]==this; }
inline void pushup() { s=1+ch[0]->s+ch[1]->s; }
}*root;
inline void rot(node* r) {
node* fa=r->fa; bool d=r->d();
fa->fa->setc(r, fa->d());
fa->setc(r->ch[!d], d);
r->setc(fa, !d);
fa->pushup();
if(fa==root) root=r;
}
inline void splay(node* r, node* fa) {
while(r->fa!=fa)
if(r->fa->fa==fa) rot(r);
else r->d()==r->fa->d()?(rot(r->fa), rot(r)):(rot(r), rot(r));
r->pushup();
}
inline node* select(node* r, const int &k) {
if(r==null) return null;
int s=r->ch[0]->s+1;
if(s==k) return r;
if(s>k) return select(r->ch[0], k);
else return select(r->ch[1], k-s);
}
void getans(node* r) {
if(r==null) return;
getans(r->ch[0]);
printf("%d ", r->key);
getans(r->ch[1]);
delete r;
}
int main() {
null=new node(0, 0); null->ch[0]=null->ch[1]=null->fa=null;
int pos, id;
while(~scanf("%d", &n)) {
root=null;
rep(i, n) {
read(pos); read(id);
++pos;
node* c=new node(id);
if(root==null) root=c;
else if(pos>root->s) {
splay(select(root, root->s), null);
root->setc(c, 1);
splay(c, null);
}
else {
splay(select(root, pos), null);
c->setc(root->ch[0], 0);
c->pushup();
root->setc(c, 0);
splay(c, null);
}
}
getans(root);
puts("");
}
return 0;
}
线段树：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
#define lc x<<1
#define rc x<<1|1
#define lson l, m, lc
#define rson m+1, r, rc
#define MID (l+r)>>1
const int N=200005;
int n, sum[N*10], a[N], b[N], ans[N];
inline void pushup(const int &x) { sum[x]=sum[lc]+sum[rc]; }
void build(const int &l, const int &r, const int &x) {
if(l==r) { sum[x]=1; return; }
int m=MID;
build(lson); build(rson);
pushup(x);
}
void update(const int &l, const int &r, const int &x, const int &id, const int &s) {
if(l==r) { ans[l]=id; sum[x]=0; return; }
int m=MID;
if(sum[lc]>=s) update(lson, id, s);
else update(rson, id, s-sum[lc]);
pushup(x);
}
int main() {
while(~scanf("%d", &n)) {
build(1, N, 1);
rep(i, n) { read(a[i]); read(b[i]); }
for3(i, n-1, 0) update(1, N, 1, b[i], a[i]+1);
for1(i, 1, n) printf("%d ", ans[i]);
puts("");
}
return 0;
}
Description
Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…
The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.
It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!
People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.
Input
There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:
- Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
- Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
There no blank lines between test cases. Proceed to the end of input.
Output
For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.
Sample Input
4 0 77 1 51 1 33 2 69 4 0 20523 1 19243 1 3890 0 31492
Sample Output
77 33 69 51 31492 20523 3890 19243
Hint
The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.
Source
POJ Monthly--2006.05.28, Zhu, Zeyuan