hdoj 4325 Flowers 【线段树 + 离散化】【区间更新单点查询】

最新推荐文章于 2018-10-22 22:06:43 发布

笑着走完自己的路

最新推荐文章于 2018-10-22 22:06:43 发布

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分类专栏：线段树离散化

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/48555701

线段树同时被 2 个专栏收录

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Flowers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2497 Accepted Submission(s): 1239

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times.
In the next N lines, each line contains two integer S_i and T_i (1 <= S_i <= T_i <= 10^9), means i-th flower will be blooming at time [S_i, T_i].
In the next M lines, each line contains an integer T_i, means the time of i-th query.

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

Sample Input

Sample Output

Case #1:
0
Case #2:
1
2
1

题意：有n朵花，已经给出每朵花的开花时间[s, e]。现在有m次查询，对查询值x，输出在时间点x开花的个数。

思路：先离散化开花的时间区间，然后就是线段树区间更新 + 单点查询

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 200000+10
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
struct Tree
{
    int l, r;
    int len;
    int sum;
    int lazy;
};
Tree tree[MAXN<<2];
void build(int o, int l, int r)
{
    tree[o].l = l, tree[o].r = r;
    tree[o].len = r - l + 1;
    tree[o].sum = tree[o].lazy = 0;
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}
void PushDown(int o)
{
    if(tree[o].lazy)
    {
        tree[ll].lazy += tree[o].lazy;
        tree[rr].lazy += tree[o].lazy;
        tree[ll].sum += tree[o].lazy * tree[ll].len;
        tree[rr].sum += tree[o].lazy * tree[rr].len;
        tree[o].lazy = 0;
    }
}
void PushUp(int o)
{
    tree[o].sum = tree[ll].sum + tree[rr].sum;
}
void update(int L, int R, int o)
{
    if(L <= tree[o].l && R >= tree[o].r)
    {
        tree[o].lazy += 1;
        tree[o].sum += tree[o].len;
        return ;
    }
    PushDown(o);
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(R <= mid)
        update(L, R, ll);
    else if(L > mid)
        update(L, R, rr);
    else
    {
        update(L, mid, ll);
        update(mid+1, R, rr);
    }
    PushUp(o);
}
int query(int o, int pos)
{
    if(tree[o].l == tree[o].r)
        return tree[o].sum;
    PushDown(o);
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(pos <= mid)
        return query(ll, pos);
    else
        return query(rr, pos);
}
int rec[MAXN];
int s[MAXN], e[MAXN];
int Q[MAXN];
int Find(int l, int r, int val)
{
    while(r >= l)
    {
        int mid = (l + r) >> 1;
        if(rec[mid] == val)
            return mid;
        else if(rec[mid] > val)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return -1;
}
int main()
{
    int t, k = 1;
    int n, m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        int p = 1;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &s[i], &e[i]);
            rec[p++] = s[i];
            rec[p++] = e[i];
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%d", &Q[i]);
            rec[p++] = Q[i];
        }
        //离散化
        sort(rec+1, rec+p);
        int R = 2;
        for(int i = 2; i < p; i++)
            if(rec[i] != rec[i-1])
                rec[R++] = rec[i];
        sort(rec+1, rec+R);
        build(1, 1, R-1);
        for(int i = 0; i < n; i++)
        {
            int x = Find(1, R-1, s[i]);
            int y = Find(1, R-1, e[i]);
            update(x, y, 1);
        }
        printf("Case #%d:\n", k++);
        for(int i = 0; i < m; i++)
        {
            int x = Find(1, R-1, Q[i]);
            printf("%d\n", query(1, x));
        }
    }
    return 0;
}