关键是加一个标记cv：这个区间有多少个结点，已被 tag 影响。
Gorgeous Sequence
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 396 Accepted Submission(s): 78
Problem Description
There is a sequence a of
length n.
We use ai to
denote the i-th
element in this sequence. You should do the following three types of operations to this sequence.
0 x y t:
For every x≤i≤y,
we use min(ai,t) to
replace the original ai‘s
value.
1 x y:
Print the maximum value of ai that x≤i≤y.
2 x y:
Print the sum of ai that x≤i≤y.
Input
The first line of the input is a single integer T,
indicating the number of testcases.
The first line contains two integers n and m denoting
the length of the sequence and the number of operations.
The second line contains n separated
integers a1,…,an (?1≤i≤n,0≤ai<231).
Each of the following m lines
represents one operation (1≤x≤y≤n,0≤t<231).
It is guaranteed that T=100, ∑n≤1000000, ∑m≤1000000.
Output
For every operation of type 1 or 2,
print one line containing the answer to the corresponding query.
Sample Input
1 5 5 1 2 3 4 5 1 1 5 2 1 5 0 3 5 3 1 1 5 2 1 5
Sample Output
5 15 3 12 Hint Please use efficient IO method
Source
2015 Multi-University Training Contest 2
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define prt(k) cout<<#k" = "<<k<<" ";
const int N = 1000003<<2;
struct Node
{
int l, r;
int mx;
ll ss;
int tag;
int cv; /// How many nodes are infected by tag in subtree
} T[N] ;
int a[N];
#define lx x<<1
#define rx x<<1|1
#define lo o<<1
#define ro o<<1|1
/// cv：这个区间有多少个结点，已被 tag 影响。
void pushup(int x)
{
T[x].ss = T[lx].ss + T[rx].ss;
T[x].mx = max(T[lx].mx, T[rx].mx);
T[x].cv = T[lx].cv + T[rx].cv;
}
void mark(int x, int t) /// da biao ji
{
if (T[x].tag!=0 && T[x].tag <= t) return ;
T[x].tag = t;
int l = T[x].l, r = T[x].r;
if (T[x].cv != r - l + 1)
{
T[x].mx = t;
T[x].ss += 1ll * (r - l + 1 - T[x].cv) * t;
T[x].cv = r - l + 1;
}
}
void pushdown(int x)
{
if (T[x].tag==0) return;
mark(lx, T[x].tag);
mark(rx, T[x].tag);
}
void fix(int o, int t )
{
if (T[o].mx < t) return;
if (T[o].tag >= t)
{
T[o].tag = 0; ///
}
if (T[o].l==T[o].r)
{
T[o].ss = T[o].mx = T[o].tag;
T[o].cv = T[o].tag != 0;
}
else
{
pushdown(o);
fix(lo, t);
fix(ro, t);
pushup(o);
}
}
void update(int o, int t , int L, int R)
{
if (T[o].mx <= t) return;
if (R < T[o].l || T[o].r < L ) return;
if (L<=T[o].l && T[o].r<=R)
{
fix(o, t);
if (T[o].l == T[o].r)
{
T[o].ss = T[o].mx= T[o].tag = t;
T[o].cv = 1;
}
else
{
pushup(o);
}
mark(o, t);
}
else
{
pushdown(o);
update(o<<1, t, L, R);
update(o<<1|1, t, L, R);
pushup(o);
}
}
void build(int o, int l, int r)
{
T[o].l = l;
T[o].r = r;
if (l==r)
{
T[o].ss = T[o].mx = T[o].tag = a[l];
T[o].cv = 1;
return;
}
int m = (l + r) >> 1;
T[o].tag = 0;
build(lo, l, m);
build(ro, m+1, r);
pushup(o);
}
Node query(int o, int L, int R)
{
Node ret ;
if (L > T[o].r || R < T[o].l)
{
ret.mx = ret.ss = 0;
return ret;
}
if (L<=T[o].l && T[o].r <= R) return T[o];
pushdown(o);
Node a = query(lo, L, R);
Node b = query(ro, L, R);
ret.mx = max(a.mx, b.mx);
ret.ss = a.ss + b.ss;
return ret;
}
/************Read**********/
char *ch, *ch1, buf[40*1024000+5], buf1[40*1024000+5];
void read(int &x)
{
for (++ch; *ch <= 32; ++ch);
for (x = 0; '0' <= *ch; ch++) x = x * 10 + *ch - '0';
}
/**************************/
int n, m;
int main()
{
ch = buf - 1;
ch1 = buf1 - 1;
fread(buf, 1, 1000 * 35 * 1024, stdin);
int re;
read(re);
while (re--)
{
read(n);
read(m);
for (int i = 1; i <= n; i++)
read(a[i]);
build(1 , 1, n);
while (m--)
{
int t, x, y, z;
read(t);
read(x);
read(y);
if (!t)
{
read(z);
update(1, z, x, y);
}
else if (t == 1) printf("%d\n", query(1, x, y).mx);
else printf("%I64d\n", query(1, x, y).ss);
}
}
return 0;
}
/**
1
5 5
1 2 3 4 5
1 1 5
2 1 5
0 3 5 3
1 1 5
2 1 5
*/
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