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

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

21 15 1042 31 44 8146

 

Sample Output

Case #1:0Case #2:121

 

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|1using 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;}