2019 ICPC 南昌网络赛I:Yukino With Subinterval(CDQ分治)
Yukino With Subinterval
Yukino has an array a_1, a_2 \cdots a_na1,a2⋯a**n. As a tsundere girl, Yukino is fond of studying subinterval.
Today, she gives you four integers l, r, x, yl,r,x,y, and she is looking for how many different subintervals [L, R][L,R] are in the interval [l, r][l,r] that meet the following restraints:
- a_L =a_{L+1} =\cdots=a_Ra**L=a**L+1=⋯=a**R, and for any i \in [L,R], x \le a_i \le yi∈[L,R],x≤a**i≤y.
- The length of such a subinterval should be maximum under the first restraint.
Note that two subintervals [L_1,R_1] , [L_2,R_2][L1,R1],[L2,R2]are different if and only if at least one of the following formulas is true:
- L1 \cancel= L2L1=L2
- R1 \cancel= R2R1=R2
Yukino, at the same time, likes making tricks. She will choose two integers pos,vpos,v, and she will change a_{pos}apo**s to vv.
Now, you need to handle the following types of queries:
- 11 pos \ vpos v: change a_{pos}apo**s to vv
- 22 l \ r \ x \ yl r x y: print the number of legal subintervals in the interval [l, r][l,r]
Input
The first line of the input contains two integers n, m (1 \le n, m \le 2 \times 10^5)n,m(1≤n,m≤2×105) – the numbers of the array and the numbers of queries respectively.
The second line of the input contains nn integers a_i (1 \le a_i \le n)a**i(1≤a**i≤n).
For the next mm line, each containing a query in one of the following queries:
- 11 pospos vv (1 \le pos, v \le n)(1≤pos,v≤n): change a_{pos}apo**s to vv
- 22 l \ r \ x \ yl r x y (1 \le l \le r \le n) (1 \le x \le y \le n)(1≤l≤r≤n)(1≤x≤y≤n): print the number of legal subintervals in the interval [l,r][l,r]
Output
For each query of the second type, you should output the number of legal subintervals in the interval [l, r][l,r].
样例输入复制
6 3
3 3 1 5 6 5
2 2 3 4 5
1 3 2
2 1 6 1 5
样例输出复制
0
4
样例解释
For the first operations, there are 33 different subintervals ([2, 2],[3, 3],[2,3])([2,2],[3,3],[2,3]) in the interval [2, 3][2,3], but none of them meets all the restraints.
For the third operations, the legal subintervals in interval [1, 6][1,6] are: [1, 2], [3, 3], [4, 4], [6, 6][1,2],[3,3],[4,4],[6,6]
Notes that although subintervals [1,1][1,1] and [2,2][2,2] also meet the first restraint, we can extend them to subinterval [1, 2][1,2]. So the length of them is not long enough, which against the second one.
题目链接：https://nanti.jisuanke.com/t/41356
题意：
给你一个含有n个数的数组，和m个操作
操作1：
将a[pos] 变为val
操作2：
询问在\([l,r]\) 中有多少个子区间满足数值在$[x,y] $ 之间 ，每一个子区间是长度尽可能大的相同数字。
思路：
将数组中 $a[i] $ ，转为在二维坐标平面上的点\((i,a[i])\) ,
那么就转为了一个带修改的二维平面中询问矩阵内点权值和的问题。
这是一个经典的三维偏序问题。
可以用CDQ分治来解决。
不会的话可以先学习一下这题：
https://www.cnblogs.com/qieqiemin/p/11613573.html
本题还需要注意几点：
因为连续相同的数值只贡献一个，所以我们把连续相同的只把第一个点放入平面中（即放入cdq的离线操作中）
那么对于每一个询问，我们就要特判一下\((l,a[l])\) 这个点，如果\(a[l]\) 在 \([x,y]\) 之间，并且 满足
$l >1 $
$a[l]==a[l-1] $
这2个条件，都需要对这个询问的答案加上1。即加上a[l]为开头的子区间的贡献，
以及修改操作，需要判断改变\(a[i]\) 对\(a[i-1]，a[i+1]\) 的影响，以及如果更改前的\(a[i]\) 是一个子区间的开头，需要去掉原来的影响（加上相反的值即可。）
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll tree[maxn];
int lowbit(int x)
{
return -x & x;
}
ll ask(int x)
{
// cout<<x<<" ";
ll res = 0ll;
while (x) {
res += tree[x];
x -= lowbit(x);
}
// cout<<res<<endl;
return res;
}
void add(int x, ll val)
{
// cout<<x<<" "<<val<<endl;
while (x < maxn) {
tree[x] += val;
x += lowbit(x);
}
}
int n, m;
struct node {
int time;
int op;
int x, y;
int val;
int ansid;
node() {}
node(int tt, int oo, int xx, int yy, int vv, int aa)
{
time = tt;
op = oo;
x = xx;
y = yy;
val = vv;
ansid = aa;
}
bool operator < (const node &bb) const
{
if (time != bb.time) {
return time < bb.time;
} else if (x != bb.x) {
return x < bb.x;
} else if (y != bb.y) {
return y < bb.y;
} else {
return op < bb.op;
}
}
bool operator<= (const node &bb )const
{
if (x != bb.x) {
return x < bb.x;
} else if (y != bb.y) {
return y < bb.y;
} else {
return op < bb.op;
}
}
} a[maxn], b[maxn];
int ans[maxn];
int tot;
int anstot;
int c[maxn];
int sym[maxn];
void cdq(int l, int r)
{
if (l == r) {
return ;
}
int mid = (l + r) >> 1;
cdq(l, mid);
cdq(mid + 1, r);
int ql = l;
int qr = mid + 1;
repd(i, l, r) {
if (qr > r || (ql <= mid && a[ql] <= a[qr])) {
if (a[ql].op == 1) {
add(a[ql].y, a[ql].val);
sym[i] = 1;
}
b[i] = a[ql++];
} else {
if (a[qr].op == 2) {
ans[a[qr].ansid] += a[qr].val * ask(a[qr].y);
}
b[i] = a[qr++];
}
}
ql = l;
qr = mid + 1;
repd(i, l, r) {if (qr > r || (ql <= mid && a[ql] <= a[qr])) {if (a[ql].op == 1) {add(a[ql].y, -a[ql].val);} ql++;} else {qr++;}}
repd(i, l, r) {a[i] = b[i];}
}
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
du2(n, m);
repd(i, 1, n) {
du1(c[i]);
if (c[i] != c[i - 1]) {
a[++tot] = node(-1, 1, i, c[i], 1, 0);
}
}
// node(int tt, int oo, int xx, int yy, int vv, int aa)
repd(i, 1, m) {
int op;
du1(op);
if (op == 1) {
int pos, v;
du2(pos, v);
if (pos != n && c[pos] == c[pos + 1]) {
a[++tot] = node(i, 1, pos + 1, c[pos + 1], 1, 0);
}
if (pos == 1 || c[pos] != c[pos - 1]) {
a[++tot] = node(i, 1, pos, c[pos], -1, 0);
}
c[pos] = v;
if (pos != n && c[pos] == c[pos + 1]) {
a[++tot] = node(i, 1, pos + 1, c[pos + 1], -1, 0);
}
if (pos == 1 || c[pos] != c[pos - 1]) {
a[++tot] = node(i, 1, pos, c[pos], 1, 0);
}
} else {
int l, r, x, y;
du2(l, r);
du2(x, y);
if (l != 1 && c[l] == c[l - 1] && c[l] <= y && c[l] >= x) {
ans[anstot]++;
}
a[++tot] = node(i, 2, r, y, 1, anstot);
a[++tot] = node(i, 2, l - 1, x - 1, 1, anstot);
a[++tot] = node(i, 2, r, x - 1, -1, anstot);
a[++tot] = node(i, 2, l - 1, y, -1, anstot++);
}
}
sort(a + 1, a + 1 + tot);
cdq(1, tot);
repd(i, 0, anstot - 1) {
printf("%d\n", max(ans[i], 0));
}
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
