FZU 1656 How many different numbers

FZU_1656

    这个题目和HDU_3333几乎一模一样，具体的思路可以参考我的HDU_3333的题解：http://www.cnblogs.com/staginner/archive/2012/04/13/2445104.html。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 100010
#define MAXQ 1010
int sum[4 * MAXD], N, Q, a[MAXD], tx[MAXD], X, where[MAXD], r[MAXQ];
struct Question
{
    int x, y, ans;
}question[MAXQ];
int cmpint(const void *_p, const void *_q)
{
    int *p = (int *)_p, *q = (int *)_q;
    return *p < *q ? -1 : 1;
}
int cmpq(const void *_p, const void *_q)
{
    int *p = (int *)_p, *q = (int *)_q;
    return question[*p].y < question[*q].y ? -1 : 1;
}
void update(int cur)
{
    int ls = cur << 1, rs = (cur << 1) | 1;
    sum[cur] = sum[ls] + sum[rs];
}
void build(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    sum[cur] = 0;
    if(x == y)
        return ;
    build(ls, x, mid);
    build(rs, mid + 1, y);
}
void init()
{
    int i, j, k;
    for(i = 0; i < N; i ++)
    {
        scanf("%d", &a[i]);
        tx[i] = a[i];
    }
    qsort(tx, N, sizeof(tx[0]), cmpint);
    X = 0;
    for(i = 0; i < N; i ++)
        if(i == 0 || tx[i] != tx[i - 1])
        {
            where[X] = -1;
            tx[X ++] = tx[i];
        }
    build(1, 0, N - 1);
    scanf("%d", &Q);
    for(i = 0; i < Q; i ++)
    {
        scanf("%d%d", &question[i].x, &question[i].y);
        -- question[i].x, -- question[i].y;
        r[i] = i;
    }
    qsort(r, Q, sizeof(r[0]), cmpq);
}
int BS(int x)
{
    int min = 0, max = X, mid;
    for(;;)
    {
        mid = (min + max) >> 1;
        if(mid == min)
            break;
        if(tx[mid] <= x)
            min = mid;
        else
            max = mid;
    }
    return mid;
}
void refresh(int cur, int x, int y, int k, int c)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x == y)
    {
        sum[cur] = c ? 1 : 0;
        return ;
    }
    if(k <= mid)
        refresh(ls, x, mid, k, c);
    else
        refresh(rs, mid + 1, y, k, c);
    update(cur);
}
int query(int cur, int x, int y, int s, int t)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
        return sum[cur];
    if(mid >= t)
        return query(ls, x, mid, s, t);
    else if(mid + 1 <= s)
        return query(rs, mid + 1, y, s, t);
    else
        return query(ls, x, mid, s, t) + query(rs, mid + 1, y, s, t);
}
void solve()
{
    int i, j, k;
    for(i = j = 0; i < N; i ++)
    {
        k = BS(a[i]);
        if(where[k] != -1)
            refresh(1, 0, N - 1, where[k], 0);
        where[k] = i;
        refresh(1, 0, N - 1, i, 1);
        while(j < Q && question[r[j]].y == i)
        {
            question[r[j]].ans = query(1, 0, N - 1, question[r[j]].x, i);
            ++ j;
        }
    }
    for(i = 0; i < Q; i ++)
        printf("%d\n", question[i].ans);
}
int main()
{
    while(scanf("%d", &N) == 1)
    {
        init();
        solve();
    }
    return 0;
}