意甲冠军：

特定n多头排列。m操作

以下是各点的颜色

以下m一种操纵：

1 l r col 染色

2 l r col 问间隔col色点

== 通的操作+区间内最大最小颜色数的优化，感觉非常不科学。。。

==感觉能够卡掉这样的写法。。反正就是不科学嘛

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
#define L(x) tree[x].l
#define R(x) tree[x].r
#define Len(x) tree[x].len
#define Lazy(x) tree[x].lazy
#define M(x) tree[x].minn
#define W(x) tree[x].maxx
#define Lson(x) (x<<1)
#define Rson(x) (x<<1|1)
const int N = 100010;
struct node{int l, r, len, lazy, minn, maxx;
}tree[N<<2];
int col[N];
void push_up(int id){if(Lazy(Lson(id)) == Lazy(Rson(id)))Lazy(id) = Lazy(Lson(id));else Lazy(id) = -1;M(id) = min(M(Lson(id)), M(Rson(id)));W(id) = max(W(Lson(id)), W(Rson(id)));
}
void push_down(int id){if(Lazy(id) != -1){Lazy(Lson(id)) = Lazy(Rson(id)) = Lazy(id);M(Lson(id)) = W(Lson(id)) = Lazy(id);M(Rson(id)) = W(Rson(id)) = Lazy(id);}
}
void build(int l, int r, int id){L(id) = l; R(id) = r;Len(id) = r-l+1;Lazy(id) = -1;if(l == r){Lazy(id) = col[l];W(id) = M(id) = col[l];return ;}int mid = (l+r)>>1;build(l, mid, Lson(id));build(mid+1, r, Rson(id));push_up(id);
}
void updata(int l, int r,int val, int id){if(l == L(id) && R(id) == r){Lazy(id) = val;W(id) = M(id) = val;return ;}push_down(id);int mid = (L(id) + R(id)) >>1;if(mid < l)updata(l, r, val, Rson(id));else if(r <= mid)updata(l, r, val, Lson(id));else {updata(l, mid, val, Lson(id));updata(mid+1, r, val, Rson(id));}push_up(id);
}
int query(int l, int r, int col, int id){if(!(M(id)<=col && col<=W(id))) return 0;if(Lazy(id)!=-1){if(Lazy(id) == col)return r-l+1;else return 0;}push_down(id);int mid = (L(id) + R(id)) >>1;if(mid < l)return query(l, r, col, Rson(id));else if(r <= mid)return query(l, r, col, Lson(id));elsereturn query(l, mid, col, Lson(id)) + query(mid+1, r, col, Rson(id));
}
int n, que;int main() {while (cin>>n>>que) {for(int i = 1; i <= n; i++)scanf("%d", &col[i]);build(1, n, 1);while(que--){int type, l, r, color;scanf("%d %d %d %d", &type, &l, &r, &color);l++; r++;if(type == 1)updata(l, r, color, 1);elseprintf("%d\n", query(l, r, color, 1));}}return 0;
}
/*
5 12
1 2 3 4 0
2 1 3 3
1 1 3 1
2 1 3 3
2 0 3 1
2 3 4 1
1 0 4 0
2 0 4 0
2 0 4 2000000000
1 0 0 1
1 4 4 2
2 0 4 1
2 0 4 2*/