Luogu 3939 数颜色

Posted 2020-12-29 czxingchen

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随手点开一个题。

技术分享图片

咦，这不是裸的动态开点线段树吗？写一个写一个……

Code：

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 3e5 + 5;

int n, qn, a[N];

inline void read(int &X) {
    X = 0;
    char ch = 0;
    int op = 1;
    for(; ch > ‘9‘ || ch < ‘0‘; ch = getchar())
        if(ch == ‘-‘) op = -1;
    for(; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
} 

struct Node {
    int lc, rc, sum;
};

namespace PSegT {
    int root[N], nodeCnt;
    Node s[N * 80];
    
    #define mid ((l + r) >> 1)
    
    void up(int p) {
        if(p) s[p].sum = s[s[p].lc].sum + s[s[p].rc].sum;
    }
    
    void ins(int &p, int l, int r, int x) {
        if(!p) p = ++nodeCnt;
        if(l == x && x == r) {
            s[p].sum++;
            return;
        }
        
        if(x <= mid) ins(s[p].lc, l, mid, x);
        else ins(s[p].rc, mid + 1, r, x);
        up(p);
    }
    
    void del(int &p, int l, int r, int x) {
        if(l == x && x == r) {
            s[p].sum--;
            if(s[p].sum == 0) p = 0;
            return;
        }
        if(x <= mid) del(s[p].lc, l, mid, x);
        else del(s[p].rc, mid + 1, r, x);
        up(p);
        
        if(!s[p].lc && !s[p].rc && !s[p].sum) p = 0; 
    } 
    
    int query(int p, int l, int r, int x, int y) {
        if(!p) return 0;
        if(x <= l && y >= r) return s[p].sum;
        
        int res = 0;
        if(x <= mid) res += query(s[p].lc, l, mid, x, y);
        if(y > mid) res += query(s[p].rc, mid + 1, r, x, y);
        return res;
    }
    
} using namespace PSegT;

inline void swap(int &x, int &y) {
    int t = x;
    x = y;
    y = t;
}

int main() {
    read(n), read(qn);
    nodeCnt = 0;
    for(int i = 1; i <= n; i++) {
        read(a[i]);
        ins(root[a[i]], 1, n, i);
    }
    
    for(int op; qn--; ) {
        read(op);
        if(op == 1) {
            int x, y, c;
            read(x), read(y), read(c);
            printf("%d
", query(root[c], 1, n, x, y));    
        } else {
            int x;
            read(x);
            del(root[a[x]], 1, n, x);
            del(root[a[x + 1]], 1, n, x + 1);
            swap(a[x], a[x + 1]);
            ins(root[a[x]], 1, n, x);
            ins(root[a[x + 1]], 1, n, x + 1);
        }
    }
    
    return 0;
}