poj 2828 Buy Tickets(树状数组 | 线段树)

题目链接：poj 2828 Buy Tickets

题目大意：给定N，表示有个人，给定每一个人站入的位置,以及这个人的权值，如今按队列的顺序输出每一个人的权值。

解题思路：第K大元素,非常巧妙，将人入队的顺序倒过来看，就是纯第K大问题，然后用树状数组还是线段树就都能够做了。

C++ 线段树

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 200005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

struct Node {
    int l, r, s;
    void set(int l, int r, int s) {
        this->l = l;
        this->r = r;
        this->s = s;
    }
}nd[maxn * 4];

int N, val[maxn], pos[maxn], rec[maxn];

void build (int u, int l, int r) {
    nd[u].set(l, r, r - l + 1);

    if (l == r)
        return ;
    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
}

int query (int u, int x) {
    nd[u].s--;

    if (nd[u].l == nd[u].r)
        return nd[u].l;

    if (nd[lson(u)].s >= x)
        return query(lson(u), x);
    else
        return query(rson(u), x - nd[lson(u)].s);
}

int main () {
    while (scanf("%d", &N) == 1) {
        build(1, 0, N - 1);
        for (int i = 0; i < N; i++)
            scanf("%d%d", &pos[i], &val[i]);

        for (int i = N - 1; i >= 0; i--)
            rec[query(1, pos[i] + 1)] = i;

        printf("%d", val[rec[0]]);
        for (int i = 1; i < N; i++)
            printf(" %d", val[rec[i]]);
        printf("\n");
    }
    return 0;
}

C++ 树状数组

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 200005;
#define lowbit(x) ((x)&(-x))

int N, fenw[maxn], pos[maxn], val[maxn], rec[maxn];

void add (int x, int v) {
    while (x <= N) {
        fenw[x] += v;
        x += lowbit(x);
    }
}

int find (int x) {
    int p = 0, s = 0;
    for (int i = 20; i >= 0; i--) {
        p += (1<<i);
        if (p > N || s + fenw[p] >= x)
            p -= (1<<i);
        else
            s += fenw[p];
    }
    return p + 1;
}

int main () {
    while (scanf("%d", &N) == 1) {
        memset(fenw, 0, sizeof(fenw));
        for (int i = 1; i <= N; i++) {
            add(i, 1);
            scanf("%d%d", &pos[i], &val[i]);
        }

        for (int i = N; i; i--) {
            int tmp = find(pos[i] + 1);
            rec[tmp] = i;
            add(tmp, -1);
        }

        for (int i = 1; i <= N; i++)
            printf("%d%c", val[rec[i]], i == N ? '\n' : ' ');
    }
    return 0;
}