[线段树]Skyscraper

题目描述

At the main street of Byteland, there will be built n skyscrapers, standing sequentially one next to other. If look leftside right, sequence of their height will be a1,a2,…,an.

Initially the street is empty, every skyscraper's height is 0. Hamster is the leader of the construction team. In each stage, Hamster can select a range [l,r], then the team will work on this range. Specifically, assume the height sequence is h1,h2,…,hn, then hl,hl+1,…,hr will increase by 1 during this stage. When hi=ai holds for all i∈[1,n], the project will be closed.

The plan may be changed for many times. There will be m events of 2 kinds below:

·1 l r k (1≤l≤r≤n,1≤k≤105), for all x∈[l,r], change ax to ax+k.
·2 l r (1≤l≤r≤n), assume a1,a2,…,al−1,ar+1,ar+2,…,an=0, ask for the minimum number of required stages to close the project.

输入

The first line of the input contains an integer T(1≤T≤1000), denoting the number of test cases.

In each test case, there are two integers n,m(1≤n,m≤100000) in the first line, denoting the number of skyscrapers and events.

In the second line, there are n integers a1,a2,...,an(1≤ai≤100000).

For the next m lines, each line describes an event.

It is guaranteed that ∑n≤106 and ∑m≤106.

输出

For each query event, print a single line containing an integer, denoting the answer.

样例输入 Copy

1
5 4
1 3 1 4 5
2 1 5
1 3 4 2
2 2 4
2 1 5

样例输出 Copy

7
6
6
题意：对于序列a[1~n]，有m个操作。1 l r k表示对[l,r]区间的数都加上k；2 l r表示询问使[l,r]区间完工的最小阶段数（一个阶段可以选择对一个子区间同时加1，当i点的值等于a[i]时，这点即完工）
思路：使[l,r]区间完工的最小阶段数，就是∑max{a[i]-a[i-1],0}。即当a[i]>a[i-1]时，对答案贡献a[i]-a[i-1]，否则对答案贡献0。
所以我们要维护a[1~n]的差分序列d[1~n]；对每一个询问[l,r]，求a[l]+sum{max(d[l+1~r],0)}（a[l]=sum{d[1~l]}）。线段树维护即可。
AC代码：

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;

ll n,m;
ll a[100005],d[100005];

ll val[100005*4],sum[100005*4];

void update(ll k){
  val[k]=val[k<<1]+val[k<<1|1];
  sum[k]=sum[k<<1]+sum[k<<1|1];
}
void build(ll k,ll l,ll r){
  if(l==r){
    val[k]=d[l];
    sum[k]=val[k]>0?val[k]:0;
    return;
  }
  ll mid=(l+r)>>1;
  build(k<<1,l,mid);
  build(k<<1|1,mid+1,r);
  update(k);
}
void modify(ll k,ll l,ll r,ll pos,ll v){
  if(l==r){
    val[k]+=v;
    sum[k]=val[k]>0?val[k]:0;
    return;
  }
  ll mid=(l+r)>>1;
  if(pos<=mid) modify(k<<1,l,mid,pos,v);
  else modify(k<<1|1,mid+1,r,pos,v);
  update(k);
}
ll query1(ll k,ll l,ll r,ll L,ll R){
  if(L>R) return 0;
  if(L<=l&&r<=R){
    return val[k];
  }
  ll ret=0;
  ll mid=(l+r)>>1;
  if(L<=mid) ret+=query1(k<<1,l,mid,L,R);
  if(R>mid) ret+=query1(k<<1|1,mid+1,r,L,R);
  return ret;
}
ll query2(ll k,ll l,ll r,ll L,ll R){
  if(L>R) return 0;
  if(L<=l&&r<=R){
    return sum[k];
  }
  ll ret=0;
  ll mid=(l+r)>>1;
  if(L<=mid) ret+=query2(k<<1,l,mid,L,R);
  if(R>mid) ret+=query2(k<<1|1,mid+1,r,L,R);
  return ret;
}

int main()
{
    ll _;scanf("%lld",&_);
    while(_--){
        scanf("%lld%lld",&n,&m);
        for(ll i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            d[i]=a[i]-a[i-1];
        }
        build(1,1,n);
        while(m--){
            ll op;scanf("%lld",&op);
            if(op==1){
                ll l,r,k;scanf("%lld%lld%lld",&l,&r,&k);
                modify(1,1,n,l,k);
                if(r+1<=n) modify(1,1,n,r+1,-k);
            }
            else{

                ll l,r;scanf("%lld%lld",&l,&r);
                ll ans=query1(1,1,n,1,l)+query2(1,1,n,l+1,r);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}