HDU 3874 Necklace

最新推荐文章于 2020-03-05 15:29:58 发布

转载最新推荐文章于 2020-03-05 15:29:58 发布 · 64 阅读

本文详细介绍了HDU_3874问题的解题策略和代码实现，包括数据结构初始化、算法构建、二分查找、更新操作、刷新和查询过程。通过实例分析，为读者提供了深入理解并解决类似问题的方法。

HDU_3874

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 50010
#define MAXQ 200010
long long int sum[4 * MAXD];
int N, Q, a[MAXD], tx[MAXD], X, r[MAXQ], where[MAXD];
struct Question
{
    int x, y;
    long long int ans;
}question[MAXQ];
int cmpint(const void *_p, const void *_q)
{
    int *p = (int *)_p, *q = (int *)_q;
    return *p - *q;
}
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 init()
{
    int i, j, k;
    scanf("%d", &N);
    for(i = 1; i <= N; i ++)
    {
        scanf("%d", &a[i]);
        tx[i] = a[i];
    }
    qsort(tx + 1, N, sizeof(tx[0]), cmpint);
    X = 0;
    for(i = 1; i <= N; i ++)
        if(i == 1 || tx[i] != tx[i - 1])
        {
            where[X] = -1;
            tx[X ++] = tx[i];
        }
    scanf("%d", &Q);
    for(i = 0; i < Q; i ++)
    {
        scanf("%d%d", &question[i].x, &question[i].y);
        r[i] = i;
    }
    qsort(r, Q, sizeof(r[0]), cmpq);
}
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);
}
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 update(int cur)
{
    sum[cur] = sum[cur << 1] + sum[(cur << 1) | 1];
}
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 ? a[x] : 0;
        return ;
    }
    if(k <= mid)
        refresh(ls, x, mid, k, c);
    else
        refresh(rs, mid + 1, y, k, c);
    update(cur);
}
long long 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;
    build(1, 1, N);
    for(i = 1, j = 0; i <= N; i ++)
    {
        k = BS(a[i]);
        if(where[k] != -1)
            refresh(1, 1, N, where[k], 0);
        where[k] = i;
        refresh(1, 1, N, i, 1);
        while(j < Q && question[r[j]].y == i)
        {
            question[r[j]].ans = query(1, 1, N, question[r[j]].x, question[r[j]].y);
            ++ j;
        }
    }
    for(i = 0; i < Q; i ++)
        printf("%I64d\n", question[i].ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;
}