codechef Many Lists(树状数组 set)

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 = 1e5 + 10, inf = 2147483647;
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]; 
    //printf("%d\n", s.size());
    sit it = s.lower_bound(mp(l, r));
    if(it != s.begin()) it--;
//  printf("%d %d\n", it -> fi, it -> se);
    while((it -> se >= l && it -> se <= r) || (it -> fi >= l && it -> se <= r)) {
        printf("%d %d\n", it -> fi, it -> se);
        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() {
    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
*/