Vases and Flowers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 4375 Accepted Submission(s): 1797
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).
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.
Output one blank line after each test case.
Sample Input
210 51 3 52 4 51 1 82 3 61 8 810 61 2 52 3 41 0 82 2 51 4 41 2 3
Sample Output
3 721 9
4Can not put any one.2 620 944 52 3
Author
SYSU
Source
Recommend
zhuyuanchen520
题解:
操作1 : 1 A F
从A开始找出为0的花瓶并将F个蝴蝶挨个放进去,输出第一个放进去的花瓶,和最后一个放进去的花瓶
二分查询一个放进去的花瓶 fir (即第一个为0的花瓶,如果fir = n + 1 表示没有空的花瓶,输出Can not put any one.)
二分查询第一个使得空花瓶数达到F的花瓶位置 las
输出fir 和 las
并将 [fir,las] 覆盖1
操作2: 2 A B
输出 [A,B]的区间和
并将[A,B] 覆盖0
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e4+10;
typedef long long ll;
struct SegTree
{
int sum[maxn<<2],lz[maxn<<2],cnt[maxn<<2];
void init() {
memset(sum,0,sizeof(sum));
memset(lz,-1,sizeof(lz));
memset(cnt,0,sizeof(cnt));
}
void push_up(int rt) {
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
cnt[rt] = cnt[rt<<1] + cnt[rt<<1|1];
}
void push_down(int rt) {
if(lz[rt] == -1) return;
lz[rt<<1] = lz[rt<<1|1] = lz[rt];
sum[rt<<1] = lz[rt] * cnt[rt<<1];
sum[rt<<1|1] = lz[rt] * cnt[rt<<1|1];
lz[rt] = -1;
}
void build(int l,int r,int rt) {
if(l == r) {
cnt[rt] = 1;
return;
}
int mid = (l+r)>>1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
push_up(rt);
}
void update(int ql,int qr,int val,int l,int r,int rt) {
if(ql == l && qr == r) {
sum[rt] = val*cnt[rt];
lz[rt] = val;
return;
}
push_down(rt);
int mid = (l+r)>>1;
if(qr <= mid) update(ql,qr,val,l,mid,rt<<1);
else if(ql > mid) update(ql,qr,val,mid+1,r,rt<<1|1);
else {
update(ql,mid,val,l,mid,rt<<1);
update(mid+1,qr,val,mid+1,r,rt<<1|1);
}
push_up(rt);
}
int query(int ql,int qr,int l,int r,int rt) {
if(ql == l && qr == r) {
return sum[rt];
}
push_down(rt);
int mid = (l+r)>>1;
if(qr <= mid) return query(ql,qr,l,mid,rt<<1);
if(ql > mid) return query(ql,qr,mid+1,r,rt<<1|1);
return query(ql,mid,l,mid,rt<<1) + query(mid+1,qr,mid+1,r,rt<<1|1);
}
}seg;
int n,m;
bool check(int l,int mid,int tot)
{
int now = seg.query(l,mid,1,n,1);
int cha = mid - l + 1 - now;
return cha >= tot;
}
int main()
{
int caset;scanf("%d",&caset);
while(caset--)
{
scanf("%d%d",&n,&m);
seg.init();
seg.build(1,n,1);
while(m--)
{
int op,sta,tot;
scanf("%d%d%d",&op,&sta,&tot);
if(op == 1) {
int l = sta+1,r = n;
while(l <= r) {
int mid = (l + r)>>1;
if(seg.query(sta+1,mid,1,n,1) == mid - sta) l = mid + 1;
else r = mid - 1;
}
int fir = l; /// 找到第一个为0的位置
if(fir == n + 1) {
printf("Can not put any one.\n");
continue;
}
tot = min(tot,n-sta-seg.query(sta+1,n,1,n,1));
l = fir,r = n;
while(l <= r) {
int mid = (l+r)>>1;
if(check(fir,mid,tot)) r = mid - 1;
else l = mid + 1;
}
int las = l; /// 找到最后一个为0的位置 (总共tot个)
printf("%d %d\n",fir-1,las-1);
seg.update(fir,las,1,1,n,1);
}
else if(op == 2){
printf("%d\n",seg.query(sta+1,tot+1,1,n,1));
seg.update(sta+1,tot+1,0,1,n,1);
}
}
printf("\n");
}
return 0;
}