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POJ 2828 Buy Tickets (线段树+反向思维)

作者:互联网

这是一道很好的题

开始想到链表,但是在第 \(k\) 个位置插入却很烦人,因为链表查询时 \(O(n)\) 的

因为是线段树题,就想到线段树上去

正着好像很难,那么反着可以么?

反着,对于每一个操作,就相当于查找前面空出 \(a\) 个格子的位置是哪个,直接线段树上搞一下就行了

正难则反的思维还是很精妙的

#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
//#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(#x".in", "r", stdin),freopen(#x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 200010 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void print(int x) { cout << x << endl ; exit(0) ; }
void PRINT(string x) { cout << x << endl ; exit(0) ; }
void douout(double x){ printf("%lf\n", x + 0.0000000001) ; }
template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
void upd(int &a, int b) { (a += b) %= MOD ; }
void mul(int &a, int b) { a = 1ll * a * b % MOD ; }

int n ;

struct node { int ps, val ; } a[N] ;
struct Tree {
    int l, r, v, num ;
    #define ls(x) x << 1
    #define rs(x) x << 1 | 1
    #define l(x) tr[x].l
    #define r(x) tr[x].r
    #define sz(x) (tr[x].r - tr[x].l + 1)
    #define v(x) tr[x].v
    #define num(x) tr[x].num
} tr[N << 2] ;

void pushup(int x) {
    v(x) = v(ls(x)) + v(rs(x)) ;
}

void build(int x, int l, int r) {
    l(x) = l, r(x) = r ;
    if (l == r) {
        v(x) = 1 ; // 初始均为空格
        return ;
    }
    int mid = (l(x) + r(x)) >> 1 ;
    build(ls(x), l, mid) ;
    build(rs(x), mid + 1, r) ;
    pushup(x) ;
}

void modify(int x, int pos, int v) {
    if (l(x) == r(x) && v(x)) {
        num(x) = v ;
        v(x) = 0 ;
        return ;
    }
    if (pos <= v(ls(x))) modify(ls(x), pos, v) ;
    else modify(rs(x), pos - v(ls(x)), v) ;
    pushup(x) ;
}

void query(int x) {
    if (l(x) == r(x)) {
        printf("%d ", num(x)) ;
        return ;
    }
    query(ls(x)) ; query(rs(x)) ;
}

signed main(){
    while (scanf("%d", &n) != EOF) {
        rep(i, 1, n) scanf("%d%d", &a[i].ps, &a[i].val) ;
        build(1, 1, n) ;
        per(i, n, 1) {
            modify(1, a[i].ps + 1, a[i].val) ;
        }
        // 查询 [1,n] 区间中有 x+1 个空格的位置并修改
        query(1) ; enter ;
    }

    return 0 ;
}

/*
写代码时请注意:
    1.ll?数组大小,边界?数据范围?
    2.精度?
    3.特判?
    4.至少做一些
思考提醒:
    1.最大值最小->二分?
    2.可以贪心么?不行dp可以么
    3.可以优化么
    4.维护区间用什么数据结构?
    5.统计方案是用dp?模了么?
    6.逆向思维?
*/


标签:Tickets,typedef,Buy,int,void,2828,include,线段,define
来源: https://www.cnblogs.com/harryhqg/p/10535079.html