• HDU 4614 Vases and Flowers (二分+线段树)


    题解思路:

    线段树用来记录空花瓶的个数

    对于每次添加花的操作 二分查找L R

    删除花的个数用 l-r+1-区间空花瓶数

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<set>
    #include<queue>
    #include<stack>
    #include<map>
    
    #define lson l,mid,now<<1
    #define rson mid+1,r,now<<1|1
    #define ls now<<1
    #define rs now<<1|1
    #define all left,right,c
    
    #define mem(a,b) memset(a,b,sizeof(a))
    #define ll long long
    
    const int maxn=1e5+10;
    
    using namespace std;
    
    int tree[4*maxn],lazy[4*maxn];///存空花1瓶书
    
    void built(int l,int r,int now)
    {
        int mid=(l+r)>>1;
        lazy[now]=-1;
        if(l==r)
        {
            tree[now]=1;
            return ;
        }
        built(lson);
        built(rson);
        tree[now]=tree[ls]+tree[rs];
    }
    
    void pushdown(int l,int r,int now)
    {
        int mid=(l+r)>>1;
        tree[ls]=(mid-l+1)*lazy[now];
        tree[rs]=(r-mid)*lazy[now];
        lazy[ls]=lazy[rs]=lazy[now];
        lazy[now]=-1;
    }
    
    void updata(int l,int r,int now,int left,int right,int c)
    {
        int mid=(l+r)>>1;
        if(left<=l&&r<=right)
        {
            tree[now]=(r-l+1)*c;
            lazy[now]=c;
            return ;
        }
        if(lazy[now]!=-1)
            pushdown(l,r,now);
        if(left<=mid) updata(lson,all);
        if(right>mid) updata(rson,all);
        tree[now]=tree[ls]+tree[rs];
    }
    
    int query(int l,int r,int now,int left,int right)
    {
        int mid=(l+r)>>1;
        if(left<=l&&r<=right)
        {
            return tree[now];
        }
        if(lazy[now]!=-1)
            pushdown(l,r,now);
        int ans=0;
        if(left<=mid) ans+=query(lson,left,right);
        if(right>mid) ans+=query(rson,left,right);
        return ans;
        tree[now]=tree[ls]+tree[rs];
    }
    
    int bin_s(int st,int ed,int mub)
    {
        int mid,l=st,r=ed;
        while(l<=r)
        {
            mid=(l+r)>>1;
            int t=query(1,ed,1,st,mid);
            if(t>=mub)
            {
                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);
            built(1,n,1);
            for(int i=1;i<=m;i++)
            {
                int cmd,a,b;
                scanf("%d%d%d",&cmd,&a,&b);
                if(cmd==1)
                {
                    a++;
                    int t=query(1,n,1,a,n);
                    if(t==0)
                    {
                        printf("Can not put any one.
    ");
                    }
                    else
                    {
                        int L=bin_s(a,n,1);
                        int R=bin_s(L,n,min(b,t));
                        updata(1,n,1,L,R,0);
                        printf("%d %d
    ",L-1,R-1);
                    }
                }
                else
                {
                    a++,b++;
                    int ans=query(1,n,1,a,b);
                    printf("%d
    ",b-a+1-ans);
                    updata(1,n,1,a,b,1);
                }
            }
            printf("
    ");
        }
        return 0;
    }
    
  • 相关阅读:
    PAT-乙级-1034. 有理数四则运算(20)
    PAT-乙级-1033. 旧键盘打字(20)
    PAT-乙级-1032. 挖掘机技术哪家强(20)
    PAT-乙级-1031. 查验身份证(15)
    PAT-乙级-1030. *完美数列(25)
    PAT-乙级-1029. 旧键盘(20)
    PAT-乙级-1028. 人口普查(20)
    PAT-乙级-1027. 打印沙漏(20)
    PAT-乙级-1026. 程序运行时间(15)
    PAT-乙级-1025. 反转链表 (25)
  • 原文地址:https://www.cnblogs.com/minun/p/10473768.html
Copyright © 2020-2023  润新知