This question is a good trouble.
The updated with the next smallest value and the minimum value, WA for a long time.
This question is used in the analysis of the potential energy practice, and did not use CPU monitor that question in the marking process.
Have time to look at the similarities and differences between these types of tree line.
code:
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define ll long long
#define N 500006
#define lson x<<1
#define rson x<<1|1
#define inf 0x3f3f3f3f
#define I(s) freopen(s".in","r",stdin)
#define O(s) freopen(s".out","w",stdout)
#define setIO(s) I(s),O(s)
using namespace std;
int nmin[N<<2],pmin[N<<2],se[N<<2];
int ptag_a[N<<2],ntag_a[N<<2],ptag_b[N<<2],ntag_b[N<<2];
void pushup(int x)
{
nmin[x]=min(nmin[lson],nmin[rson]);
pmin [x] = min (P min [lson] pmin [rson]);
if (nmin [lson] <nmin [rson]) See [x] = min (see [lson], nmin [rson]);
if (nmin [rson] <nmin [lson tax]) [x] = min (se [rson], nmin [lson]);
if (nmin [lson] == nmin [rson tax]) [x] = min (se [lson tax], [rson]);
}
// v1表示历史最小
void mark_a (int x, int v1, int v2)
{
pmin [x] = min (pmin [x], nmin [x] + v1);
ptag_a [x] = min (ptag_a [x], ntag_a [x] + v1);
nmin [x] + = v2, ntag_a [x] + = v2;
}
Void mark_b (int x, int v1, int v2)
{
ptag_b [x] = min (ptag_b [x], ntag_b [x] + v1);
ntag_b [x] + = v2;
if (se [x]! = tax INF) [x] + = v2;
}
Void pushdown (int x)
{
int c = min (nmin [lson], nmin [rson]);
if (nmin [lson] == c)
{
}
mark_a (lson, ptag_a [x], ntag_a [x]);
mark_b (lson, ptag_b [x], ntag_b [x]);
}
Else
{
mark_a (lson, ptag_b [x], ntag_b [x]);
mark_b (lson, ptag_b [x], ntag_b [x]);
}
If (nmin [rson] == c)
{
mark_a (rson, ptag_a [x], ntag_a [x]);
mark_b (rson, ptag_b [x], ntag_b [x]);
}
Else
{
mark_a (rson, ptag_b [x], ntag_b [x]);
mark_b (rson, ptag_b [x], ntag_b [x]);
ptag_a [x] = ntag_a [x] = ptag_b [x] = ntag_b [x] = 0;
}
Void addv (int l, int r, int x, int L, R int, int v)
{
if(l>=L&&r<=R)
{
mark_a(x,v,v),mark_b(x,v,v);
return;
}
pushdown(x);
int mid=(l+r)>>1;
if(L<=mid) addv(l,mid,lson,L,R,v);
if(R>mid) addv(mid+1,r,rson,L,R,v);
pushup(x);
}
void getmax(int l,int r,int x,int L,int R,int v)
{
if(nmin[x]>=v) return;
if(l>=L&&r<=R&&se[x]>v)
{
mark_a(x,v-nmin[x],v-nmin[x]);
return;
}
pushdown(x);
int mid=(l+r)>>1;
if(L<=mid) getmax(l,mid,lson,L,R,v);
if(R>mid) getmax(mid+1,r,rson,L,R,v);
pushup(x);
}
int queryn(int l,int r,int x,int L,int R)
{
if(l>=L&&r<=R) return nmin[x];
pushdown(x);
int mid=(l+r)>>1;
if(L<=mid&&R>mid) return min(queryn(l,mid,lson,L,R),queryn(mid+1,r,rson,L,R));
else if(L<=mid) return queryn(l,mid,lson,L,R);
else return queryn(mid+1,r,rson,L,R);
}
int queryp(int l,int r,int x,int L,int R)
{
if(l>=L&&r<=R) return pmin[x];
pushdown(x);
int mid=(l+r)>>1;
if(L<=mid&&R>mid) return min(queryp(l,mid,lson,L,R),queryp(mid+1,r,rson,L,R));
else if(L<=mid) return queryp(l,mid,lson,L,R);
else return queryp(mid+1,r,rson,L,R);
}
void build(int l,int r,int x)
{
if(l==r)
{
scanf("%d",&nmin[x]),pmin[x]=nmin[x],se[x]=inf;
return;
}
int mid=(l+r)>>1;
build(l,mid,lson),build(mid+1,r,rson);
pushup(x);
}
int main()
{
// setIO("input");
int n,m;
scanf("%d%d",&n,&m),build(1,n,1);
for(int i=1;i<=m;++i)
{
int op,l,r,z;
scanf("%d%d%d",&op,&l,&r);
if(op==1) scanf("%d",&z) addv (1, n, 1, l, r, z);
if(op==2) scanf("%d",&z),getmax(1,n,1,l,r,z);
if(op==3) printf("%d\n",queryn(1,n,1,l,r));
if(op==4) printf("%d\n",queryp(1,n,1,l,r));
}
return 0;
}