Vases and Flowers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 347 Accepted Submission(s): 108
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.
http://acm.hdu.edu.cn/showproblem.php?pid=4614
思路：很基本的线段树问题，我们可以用线段树维护每一个区间的剩余空间数量（也就是可以插画的地方），对于第一个询问，我们计算区间[a,n-1]所剩余的空间数，假设为num，若为0，则输出 “Can not put any one.”，否则若小于F，则将F设为num，接下来可以二分区间[a,mid]来确定插入的第一只花和最后一只花的位置l,r，然后将区间[l,r]全赋值为1即可，对于询问2，则计算出区间[a,b]所剩余的空间num，然后 b-a+1-num即为答案。最后将区间[a,b]赋值为1.以上，下面是代码.
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#define maxn 50010
#define mid ((t[p].l+t[p].r)>>1)
#define ls (p<<1)
#define rs (ls|1)
using namespace std;
struct tree
{
int l,r;
int sum;
int lazy;
}t[maxn<<2];
void pushup(int p)
{
t[p].sum=t[ls].sum+t[rs].sum;
}
void pushdown(int p)
{
if(t[p].lazy==0)
{
t[ls].lazy=0;
t[rs].lazy=0;
t[ls].sum=0;
t[rs].sum=0;
t[p].lazy=-1;
}
else if(t[p].lazy==1)
{
t[ls].lazy=1;
t[rs].lazy=1;
t[ls].sum=t[ls].r-t[ls].l+1;
t[rs].sum=t[rs].r-t[rs].l+1;
t[p].lazy=-1;
}
}
void build(int p,int l,int r)
{
t[p].l=l,t[p].r=r,t[p].sum=1,t[p].lazy=-1;
if(l==r)
return;
build(ls,l,mid);
build(rs,mid+1,r);
pushup(p);
}
void change(int p,int l,int r,int val)
{
if(t[p].l==l&&t[p].r==r)
{
t[p].lazy=val;
if(val)
{
t[p].sum=r-l+1;
}
else
t[p].sum=0;
return;
}
pushdown(p);
if(l>mid)
change(rs,l,r,val);
else if(r<=mid)
change(ls,l,r,val);
else
{
change(ls,l,mid,val);
change(rs,mid+1,r,val);
}
pushup(p);
}
int query(int p,int l,int r)
{
if(t[p].l==l&&t[p].r==r)
{
return t[p].sum;
}
pushdown(p);
if(l>mid)
return query(rs,l,r);
else if(r<=mid)
return query(ls,l,r);
else
return query(ls,l,mid)+query(rs,mid+1,r);
}
int main()
{
//freopen("dd.txt","r",stdin);
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
int n,m,i;
scanf("%d%d",&n,&m);
build(1,0,n-1);
int k,a,b;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&k,&a,&b);
if(k==1)
{
int ttmp=query(1,a,n-1);
if(ttmp==0)
{
printf("Can not put any one.\n");
}
else
{
if(ttmp<b)
b=ttmp;
int mi=a,ma=n-1,Mid;
int l,r;
while(mi<=ma)
{
Mid=(mi+ma)>>1;
if(query(1,a,Mid)>0)
{
l=Mid;
ma=Mid-1;
}
else
mi=Mid+1;
}
mi=a,ma=n-1;
while(mi<=ma)
{
Mid=(mi+ma)>>1;
if(query(1,a,Mid)>=b)
{
r=Mid;
ma=Mid-1;
}
else
mi=Mid+1;
}
change(1,l,r,0);
printf("%d %d\n",l,r);
}
}
else
{
printf("%d\n",b-a+1-query(1,a,b));
change(1,a,b,1);
}
}
printf("\n");
}
return 0;
}
