题意

题目链接

Sol

直接做肯定不好搞(反正我不会。。)

直接开\(n\)个Pair类型的set，维护每个数的出现位置

每次在set中二分后暴力合并即可

然后就是树状数组的基本操作了

时间复杂度：\(O(nlog^2n)\)

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
#define lb(x) (x & (-x))
using namespace std;
const int MAXN = 1e6 + 10, INF = 2147483646;
inline int read() {
    char c = getchar(); int x = 0, f = 1; 
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, T[MAXN];
void Add(int x, int v) {
    while(x <= N) T[x] += v, x += lb(x);
}
void AddInt(int l, int r, int val) {
    Add(l, val); Add(r + 1, -val);
} 
int Query(int x) {
    int ans = 0;
    while(x) ans += T[x], x -= lb(x);
    return ans;
}
set<Pair> S[MAXN];
#define sit set<Pair>::iterator 
void Add(int l, int r, int x) {
    set<Pair> &s = S[x]; 
    sit it = s.lower_bound(MP(l, r));
    if(it != s.begin()) {
        it--;
        if(it -> se > r) return ;
        if(it -> se >= l ) {
            l = min(l, it -> fi); r = max(r, it -> se);
            AddInt(it -> fi, it -> se, -1);
            s.erase(it++);
        }       
    }
    it = s.lower_bound(MP(l, r));
    while((it -> se >= l && it -> se <= r) || (it -> fi >= l && it -> fi <= r)) {
        l = min(l, it -> fi); r = max(r, it -> se);
        AddInt(it -> fi, it -> se, -1);
        s.erase(it++);
    }
    s.insert(MP(l, r));
    AddInt(l, r, 1);
}
int main() {
//  freopen("a.in", "r", stdin); freopen("b.out", "w", stdout);
    N = read(); M = read();
    for(int i = 1; i <= N; i++) S[i].insert(MP(INF, INF)); 
    while(M--) {
        int opt = read();
        if(opt == 0) {int l = read(), r = read(), val = read(); Add(l, r, val);}
        else printf("%d\n", Query(read()));
    }
    return 0;
}
/*
5 5
0 2 4 4
0 3 5 5
0 1 5 5
1 5
1 3
5 12
0 1 1 1
1 1
0 1 2 1
1 1
0 1 3 1
1 1
0 2 3 1
1 1
0 4 5 1
1 1
0 1 5 1
1 1
*/