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