题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4614
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Problem Description
Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, …, N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.
Input
The first line contains an integer T, indicating the number of test cases.
For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).
Output
For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output ‘Can not put any one.’. For each operation of which K is 2, output the number of discarded flowers.
Output one blank line after each test case.
Sample Input
2
10 5
1 3 5
2 4 5
1 1 8
2 3 6
1 8 8
10 6
1 2 5
2 3 4
1 0 8
2 2 5
1 4 4
1 2 3
Sample Output
3 7
2
1 9
4
Can not put any one.
2 6
2
0 9
4
4 5
2 3
Problem solving report:
Description: 现在有N个花瓶，一开始每个花瓶都是空的，现在有两种操作：
①1 x y，表示以x号花瓶作为起点，让我们往右插y朵花，遇到空瓶子就放一朵花进去，直到花朵放完或没有瓶子。输出第一个插花的位置，和最后一个插花的位置。
②2 x y，让我们先输出从x到y号花瓶中，一共有多少朵花，然后再清空所有花瓶。
Problem solving: 对于第二种操作，我们直接区间查询然后区间更新即可。对于第一种操作：
①我们首先二分出第一个插花的位子;
②然后再二分最后一个插花的位子即可。
线段树维护的就是区间和，就是线段树模板了。
Accepted Code:
/*
* @Author: lzyws739307453
* @Language: C++
*/
#include <bits/stdc++.h>
using namespace std;
#define lson rt << 1
#define rson rt << 1 | 1
const int MAXN = 5e4 + 5;
//线段树维护值
struct Tree {
int val, laz;
}seg[MAXN << 2];
//建树
void Create(int l, int r, int rt) {
seg[rt].val = 0;
seg[rt].laz = -1;
if (!(r - l))
return ;
int mid = l + (r - l >> 1);
Create(l, mid, lson);
Create(mid + 1, r, rson);
}
//上推
void PushUp(int rt) {
seg[rt].val = seg[lson].val + seg[rson].val;
}
//下压
void PushDown(int l, int r, int rt) {
if (~seg[rt].laz) {
int mid = l + (r - l >> 1);
seg[lson].laz = seg[rson].laz = seg[rt].laz;
seg[lson].val = (mid - l + 1) * seg[rt].laz;
seg[rson].val = (r - mid) * seg[rt].laz;
seg[rt].laz = -1;
}
}
//更新
void Update(int Ql, int Qr, int v, int l, int r, int rt) {
if (Ql <= l && Qr >= r) {
seg[rt].val = (r - l + 1) * v;
seg[rt].laz = v;
return ;
}
PushDown(l, r, rt);
int mid = l + (r - l >> 1);
if (Ql <= mid)
Update(Ql, Qr, v, l, mid, lson);
if (Qr > mid)
Update(Ql, Qr, v, mid + 1, r, rson);
PushUp(rt);
}
//查询
int Query(int Ql, int Qr, int l, int r, int rt) {
if (Ql <= l && Qr >= r)
return seg[rt].val;
PushDown(l, r, rt);
int ans = 0;
int mid = l + (r - l >> 1);
if (Ql <= mid)
ans += Query(Ql, Qr, l, mid, lson);
if (Qr > mid)
ans += Query(Ql, Qr, mid + 1, r, rson);
return ans;
}
//二分
int bin_Search(int ll, int rr, int x) {
int l = ll, r = rr;
while (l <= r) {
int mid = l + (r - l >> 1);
if (mid - ll + 1 - Query(ll, mid, 1, rr, 1) >= x)
r = mid - 1;
else l = mid + 1;
}
return l;
}
int main() {
int t, n, m;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
Create(1, n, 1);
while (m--) {
int op, l, r;
scanf("%d%d%d", &op, &l, &r);
if (op != 2) {
int cnt = n - l - Query(l + 1, n, 1, n, 1);
if (!cnt)
printf("Can not put any one.\n");
else {
int ll = bin_Search(l + 1, n, 1);
int rr = bin_Search(l + 1, n, min(cnt, r));
Update(ll, rr, 1, 1, n, 1);
printf("%d %d\n", ll - 1, rr - 1);
}
}
else {
printf("%d\n", Query(l + 1, r + 1, 1, n, 1));
Update(l + 1, r + 1, 0, 1, n, 1);
}
}
printf("\n");
}
return 0;
}
