HDU-4614 Vases and Flowers（线段树区间更新+二分查找）

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

题目大意：
有n个花瓶，标号0 ~ n−1。有m个操作，

‘1 A F′，表示从A位置开始插F朵花，遇到有花的花瓶跳过。到最后一个花瓶都还有花剩余，丢弃剩下的花。

‘2 A B′，表示将区间[A,B]内的花瓶全部清空。(A≤B) 

对于每个1操作，输出第一个和最后一个插花的位置，如果一朵花都插不了，输出‘Can not put any one.’；对于每个2操作，输出区间[A,B]内被清空的花瓶的数量。

解题思路：

直接线段树存区间内的空花瓶数量。原来标号0 ~ n−1，但我线段树序号习惯了1 ~ n。

设num(l,r)为区间[l,r]的空花瓶数。

对于一个1操作，首先判一下num(A,n)是否大于0。

因为区间[A,n]的空花瓶数是单调不递减的，所以可以通过二分查找到 一个最小的位置L，使得num(A,L)=1，则此时的L就是第一个插花的位置。

同样二分找到一个最小的位置R，使得num(A,R)=min(F,num(A,n))，则此时的R就是最后一个插花的位置。（输出时记得减1）

对于一个2操作，直接询问区间[A,B]的空花瓶数，拿总数一减，输出，然后更新区间即可。

代码如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <math.h>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e5+10;
using namespace std;

struct SegTree_node
{
    int l;
    int r;
    int num;//区间内空瓶的数量
    int tag;//如果为-1则为初始状态,不用向下更新，如果为0意思将子区间内的花清空，为1为插满花
}SegTree[50005<<2]; 

int n,m;

void PushUp(int rt)//向上传递区间信息
{
    SegTree[rt].num=SegTree[rt<<1].num+SegTree[rt<<1|1].num;
}

void PushDown(int rt)//向下传递状态
{
    if(SegTree[rt].tag!=-1)
    {
        SegTree[rt<<1].tag=SegTree[rt<<1|1].tag=SegTree[rt].tag;
        int mid=(SegTree[rt].l+SegTree[rt].r)>>1;
        SegTree[rt<<1].num=(mid-SegTree[rt].l+1)*SegTree[rt].tag;
        SegTree[rt<<1|1].num=(SegTree[rt].r-mid)*SegTree[rt].tag;
        SegTree[rt].tag=-1;
    }
}

void Build(int l,int r,int rt)
{
    SegTree[rt].l=l;
    SegTree[rt].r=r;
    SegTree[rt].num=r-l+1;
    SegTree[rt].tag=-1;
    if(l==r)
        return ;
    int mid=(l+r)>>1;
    Build(l,mid,rt<<1);
    Build(mid+1,r,rt<<1|1);
}

void Update(int L,int R,int f,int rt)
{
    int l=SegTree[rt].l;
    int r=SegTree[rt].r;
    if(L<=l&&R>=r)
    {
        SegTree[rt].num=(r-l+1)*f;//先计算 
        SegTree[rt].tag=f;//标记当前区间
        return ;
    }
    if(l==r)//不要忘记 
        return ;
    PushDown(rt);//向下传递状态
    int mid=(l+r)>>1;
    if(L<=mid)
        Update(L,R,f,rt<<1);
    if(R>mid)
        Update(L,R,f,rt<<1|1);
    PushUp(rt);
}

int Query(int L,int R,int rt)//得到区间[L,R]中的空花瓶数
{
    int l=SegTree[rt].l;
    int r=SegTree[rt].r;
    if(R<l||L>r)
        return 0;
    if(L<=l&&R>=r)
        return SegTree[rt].num;
    PushDown(rt);
    int mid=(l+r)>>1;
    int sum=0;
    if(L<=mid)
        sum+=Query(L,R,rt<<1);
    if(R>mid)
        sum+=Query(L,R,rt<<1|1);
    return sum;
}

int dive(int x,int num)//二分查找，第一个为开始的位置，第二个参数为要插花的个数 
{
    int l=x;
    int r=n;
    int ans=0;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(Query(x,mid,1)>=num)
        {
            ans=mid;
            r=mid-1;
        }
        else
            l=mid+1;
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        Build(1,n,1);
        for(int i=1;i<=m;i++)
        {
            int op;
            scanf("%d",&op);
            if(op==1)
            {
                int A,F;
                scanf("%d %d",&A,&F);
                A++;//存储时序号从一开始，故序号要全加一，下同 
                int cnt=Query(A,n,1);//得到区间[A, n]中的空花瓶数
                if(cnt==0)
                    printf("Can not put any one.
");
                else
                {
                    int L=dive(A,1);//二分左端点(第1个能插花的位置)
                    int R=dive(A,min(cnt,F));//二分右端点(最后一个能插花的位置)
                    Update(L,R,0,1);//将区间[L,R]的花瓶全部插满
                    printf("%d %d
",L-1,R-1);
                }
            }
            else if(op==2)
            {
                int A,B;
                scanf("%d %d",&A,&B);
                A++;
                B++;
                printf("%d
",B-A+1-Query(A,B,1));
                Update(A,B,1,1);//将区间[A,B]的花瓶全部清空
            }
        }
        printf("
");
    } 
    return 0;
}

下面是kuangbin大佬用另一种模型的线段树写的，虽然有点麻烦，但是有空还是值得一看：https://www.cnblogs.com/kuangbin/p/3214805.html