POJ 2828Buy Tickets

POJ 2828

题目大意是说有n个插入操作，每次把B插入到位置A，原来A以后的全部往后移动1,球最后的序列

tree里保存的应该是这整个区间还有多扫个位置可以插入数据，那么线段树里从后往前扫描依次插入数据

比如现在吧B插入到A位置，如果整个区间左侧还有<A个位置可以插入数据，那么就只能将其放到整个区间的右侧，递归下去就可以了

void update(int k, int L, int R, int x)
{
    if(L == R) { pre[k] = r; ans[L] = val; return ; }

    int mid = (L+R)>>1;

    if(x <= pre[k<<1]) update(lson, x);//左侧的多余x

    else update(rson, x-pre[k<<1]);//否则往右

     pre[k] = pre[k<<1] + pre[k<<1|1];
}

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 1e9
#define inf (-((LL)1<<40))
#define lson k<<1, L, mid
#define rson k<<1|1, mid+1, R
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define FOPENIN(IN) freopen(IN, "r", stdin)
#define FOPENOUT(OUT) freopen(OUT, "w", stdout)
template<class T> T CMP_MIN(T a, T b) { return a < b; }
template<class T> T CMP_MAX(T a, T b) { return a > b; }
template<class T> T MAX(T a, T b) { return a > b ? a : b; }
template<class T> T MIN(T a, T b) { return a < b ? a : b; }
template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

typedef __int64 LL;
//typedef long long LL;
const int MAXN = 200005;
const int MAXM = 100005;
const double eps = 1e-10;
const LL MOD = 1000000007;

int N, pre[MAXN<<2];
int id[MAXN], num[MAXN], ans[MAXN];
int r, val;

int buildTree(int k, int L, int R)
{
    if(L == R) return pre[k] = 1;
    int mid = (L+R)>>1;
    return pre[k] = buildTree(lson) + buildTree(rson);
}

void update(int k, int L, int R, int x)
{
    if(L == R) { pre[k] = r; ans[L] = val; return ; }

    int mid = (L+R)>>1;

    if(x <= pre[k<<1]) update(lson, x);

    else update(rson, x-pre[k<<1]);

     pre[k] = pre[k<<1] + pre[k<<1|1];
}

int main()
{
    while(~scanf("%d", &N))
    {
        buildTree(1, 1, N);
        for(int l=1;l<=N;l++)
            scanf("%d %d", &id[l], &num[l]);
        r = 0;
        for(int i=N;i>0;i--)
        {
            val = num[i];
            update(1, 1, N, id[i] + 1);
        }
        for(int i=1;i<=N;i++) printf("%d%c", ans[i], i==N?'
':' ');
    }
    return 0;
}