CodeForces 52C Circular RMQ （线段树的区间更新+lazy tag）-蒲公英云

CodeForces 52C Circular RMQ （线段树的区间更新+lazy tag）

忘是亡心i 2022-06-12 02:44 245阅读 0赞

You are given circular array a0, a1, …, a**n - 1. There are two types of operations with it:

inc(lf, rg, v) — this operation increases each element on the segment [lf, rg](inclusively) by v;
rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg](inclusively).

Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.

Write program to process given sequence of operations.

Input

The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, …, a**n - 1 ( - 106 ≤ a**i ≤ 106), a**i are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next mlines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation.

Output

For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Example

Input

Output

1
0
0

题解：

现在还在比赛。。但我已经写不出来题了，就来写博客了，好高兴线段树的题没看题解就写出来了，这几天线段树和树状数组的题没白刷，这题就是如果输入区间不是从小到大就分成两个区间做，然后要注意一下区间更新要lazy tag，延迟更新，然后输入的时候做点判断处理这题就ac了

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<deque>
using namespace std;
struct node
{
    int l,r;
    long long minn;
    long long tag;
};
node t[200005*4];
int a[200005];
void Build(int l,int r,int num)//日常建树
{
    t[num].l=l;
    t[num].r=r;
    t[num].tag=0;
    if(l==r)
    {
        t[num].minn=a[l];
        return;
    }
    int mid=(l+r)/2;
    Build(l,mid,num*2);
    Build(mid+1,r,num*2+1);
    t[num].minn=min(t[num*2].minn,t[num*2+1].minn);//更新下最小值
}
void update(int l,int r,int num,int v)
{
    if(t[num].l==l&&t[num].r==r)
    {
        t[num].minn+=v;
        t[num].tag+=v;
        return;
    }
    if(t[num].tag!=0)//除了这个都是日常更新，这个可以写成一个函数lazy tag，将上次没有更新的东西向下子节点更新
    {
        t[num*2].minn+=t[num].tag;
        t[num*2+1].minn+=t[num].tag;
        t[num*2].tag+=t[num].tag;
        t[num*2+1].tag+=t[num].tag;
        t[num].tag=0;//清除状态
    }
    int mid=(t[num].l+t[num].r)/2;
    if(r<=mid)
        update(l,r,num*2,v);
    else if(l>mid)
        update(l,r,num*2+1,v);
    else
    {
        update(l,mid,num*2,v);
        update(mid+1,r,num*2+1,v);
    }
    t[num].minn=min(t[num*2].minn,t[num*2+1].minn);//更新最小值
}
long long query(int l,int r,int num)//访问
{
    if(t[num].l==l&&t[num].r==r)
    {
        return t[num].minn;
    }
      if(t[num].tag!=0)//和上面一样，pushdown
    {
        t[num*2].minn+=t[num].tag;
        t[num*2+1].minn+=t[num].tag;
        t[num*2].tag+=t[num].tag;
        t[num*2+1].tag+=t[num].tag;
        t[num].tag=0;
    }
    int mid=(t[num].l+t[num].r)/2;
    if(r<=mid)
        return query(l,r,num*2);
    else if(l>mid)
    {
        return query(l,r,num*2+1);
    }
    else
        return min(query(l,mid,num*2),query(mid+1,r,num*2+1));
}
int main()
{
    int i,j,n,m,x,y,z,tag,flag;
    char s[100];
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    Build(0,n-1,1);
    scanf("%d",&m);
    getchar();//吸收回车
    for(i=0;i<m;i++)
    {
        tag=0;
        gets(s);
        x=0,y=0,z=0;
        flag=0;
        for(j=0;j<strlen(s);j++)
        {
            if(s[j]==' ')//根据空格数目判断情况，储存数据
            {
                tag++;
                continue;
            }
            if(tag==0)
            {
                x=x*10+s[j]-'0';
            }
            else if(tag==1)
            {
                y=y*10+s[j]-'0';
            }
            else
            {
                if(s[j]=='-')
                    flag=1;
                else
                {
                    z=z*10+s[j]-'0';
                }
            }
        }
        if(flag)//z处有负号
            z=-z;
        if(tag==1)
        {
            if(x<=y)
                printf("%I64d\n",query(x,y,1));//codeforce要用i64d
            else
                printf("%I64d\n",min(query(0,y,1),query(x,n-1,1)));//如果输入的y大于x，就分成两个部分
        }
        else
        {
            if(x<=y)
                update(x,y,1,z);
            else
            {
                update(0,y,1,z);//同理分成两个部分
                update(x,n-1,1,z);
            }
        }
    }
    return 0;
}