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自然万物都趋向从有序变得无序

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Circular RMQ CodeForces - 52C(线段树)

发表于 更新于 分类于 Valine：

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题目

You are given circular array a0, a1, …, an - 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, …, an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines 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

4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1

Output

1
0
0

题意:

给你一个循环序列,有两种操作:

  1. 区间(lg, rg) 的所有元素加上V;
  2. 查询区间(lg, rg) 的最小值.

思路:

线段树的区间查询与更新,我们用线段树维护区间的最小值,并且设置一个懒惰标记来延迟更新.

代码:

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;

typedef long long LL;
const LL INF = 0x3fffffff;
struct ss{
    LL v,tag;//tag是懒惰标记
}a[200000 << 2];
LL Min(LL a,LL b){
    return a < b ? a:b;
}

void pushup(int cur){//向上更新
    a[cur].v = Min(a[cur<<1].v,a[cur<<1|1].v);
}

void built(int l,int r,int cur){
    a[cur].v = 0;
    a[cur].tag = 0;//0 表示没有被标记
    if(l == r){
        scanf("%I64d", &a[cur].v);
        return ;
    }
    int mid = (l + r) >> 1;
    built(l, mid,cur << 1);
    built(mid + 1,r,cur << 1|1);
    pushup(cur);
}

void pushdown(int cur){
    if(a[cur].tag != 0){//向下调整,将标记传给儿子
        a[cur<<1].tag += a[cur].tag;
        a[cur<<1|1].tag += a[cur].tag;
        a[cur<<1].v += a[cur].tag;
        a[cur<<1|1].v += a[cur].tag;
        a[cur].tag = 0;
    }
}

void update(int l,int r,int ll,int rr,int cur,LL v){
    if(l<=ll&&r>=rr){
        a[cur].v += v;
       a[cur].tag += v;
        return ;
    }pushdown(cur);
    int mid = (ll +rr)>>1;
    if(mid < r) update(l,r,mid + 1,rr,cur<<1|1,v);
    if(mid >= l) update(l,r,ll,mid,cur<<1,v);
    pushup(cur);
}

LL quiry(int l,int r,int ll,int rr,int cur){
    if(l<=ll&&r>=rr){
        return a[cur].v;
    } pushdown(cur);
    int mid = (ll + rr) >> 1;
    LL t1 =INF;
    LL t2 = INF;
    if(mid < r) t1 = quiry(l,r,mid + 1,rr,cur<<1|1);
    if(mid >= l) t2 =  quiry(l,r,ll,mid,cur<<1);
    return Min(t1,t2);
}

int main(){
    int n;
    scanf("%d", &n);
    built(1,n,1);
    int m;
    scanf("%d", &m);
    while(m--){
        int a,b,c;
        char ch;
        scanf("%d%d%c", &a, &b, &ch);
        a++,b++;
      if(ch != ' '){
        if(a>b){
            LL t1=quiry(a,n,1,n,1);
            LL t2 = quiry(1,b,1,n,1);
            printf("%I64d\n",Min(t1,t2));
        }else
            printf("%I64d\n",quiry(a,b,1,n,1));
      }else {scanf("%d", &c);
        if(a>b){
            update(a,n,1,n,1,c);
            update(1,b,1,n,1,c);
        }else update(a,b,1,n,1,c);
      }
    }
    return 0;
}