數顏色 （二分查找 主席樹）

數顏色

10.23

寫了個主席樹，結果Tdiao了。。。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define N 300010
using namespace std;

struct node{
    int sum;
    node *ls, *rs;
    void update( ){
        sum = ls->sum + rs->sum;
    }
}pool[N * 60], *tail = pool, *root[N], *null;

int n, m, maxx = 0;
int a[N];

inline void init(){
    null = ++tail;
    null->ls = null->rs = null;
    null->sum = 0;
}

inline node *newnode(){
    node* nd = ++tail;
    nd->sum = 0;
    nd->ls = nd->rs = null;
    return nd;
}

inline void insert(node *ne, node * &nd, int lf, int rg, int pos){
    nd = newnode();
    nd->sum = ne->sum + 1;
    nd->ls = ne->ls, nd->rs = ne->rs;
    if(lf == rg) return ;
    int mid = (lf + rg) >> 1;
    if(pos <= mid) insert(ne->ls, nd->ls, lf, mid, pos);
    else insert(ne->rs, nd->rs, mid+1, rg, pos);
    nd->update();
}

inline void modify(node *nd, int lf, int rg, int pos, int delta){
    node* ne = newnode();
    if(lf == rg){
        nd->sum += delta;
        return ;
    }
    int mid = (lf + rg) >> 1;
    if(pos <= mid) ne->ls = nd->ls->ls, ne->rs = nd->ls->rs, ne->sum = nd->ls->sum, nd->ls = ne, modify(nd->ls, lf, mid, pos, delta);
    else ne->ls = nd->rs->ls, ne->rs = nd->rs->rs, ne->sum = nd->rs->sum, nd->rs = ne, modify(nd->rs, mid+1, rg, pos, delta);
    nd->update();
}

inline int query(node *ne, node *nd, int lf, int rg, int pos){
    if(lf == rg) return nd->sum - ne->sum;
    int mid = (lf + rg) >> 1;
    if(pos <= mid) return query(ne->ls, nd->ls, lf, mid, pos);
    else return query(ne->rs, nd->rs, mid+1, rg, pos);
}

int main(){
    //freopen ("color2.in", "r", stdin);
    freopen ("color.in", "r", stdin);
    freopen ("color.out", "w", stdout);
    //cout << sizeof(pool) << endl;
    init();
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++) scanf("%d", &a[i]);
    int lim = 300010;
    root[0] = newnode();
    for(int i=1; i<=n; i++) insert(root[i-1], root[i], 1, lim, a[i]);
    while( m-- ){
        int opt, L, R, C;
        scanf("%d", &opt);
        if(opt == 1){
            scanf("%d%d%d", &L, &R, &C);
            printf("%d\n", query(root[L-1], root[R], 1, lim, C));
        }
        else{
            scanf("%d", &L);
            modify(root[L], 1, lim, a[L], -1);
            modify(root[L], 1, lim, a[L+1], 1);
            swap(a[L], a[L+1]);
        }
    }
    return 0;
}

粘排序，二分查找的做法吧

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#define N 300010
using namespace std;

int n, m;
int a[N];
vector <int> b[N];

int main() {
    freopen("color.in", "r", stdin);
    freopen("color.out", "w", stdout);
    scanf("%d%d", &n, &m);
    for (int i=1; i<=n; i++) {
        scanf("%d", &a[i]);
        b[a[i]].push_back(i);
    }
    for (int i=1; i<=n; i++) sort(b[i].begin(), b[i].end());
    for (int i=1, x, l, r, c; i<=m; i++) {
        scanf("%d", &x);
        if (x == 1) {
            scanf("%d%d%d", &l, &r, &c);
            printf("%d\n", (int)(upper_bound(b[c].begin(), b[c].end(), r) - lower_bound(b[c].begin(), b[c].end(), l)));
        }
        else {
            scanf("%d", &x);
            if (a[x] != a[x + 1]) {
                (*lower_bound(b[a[x]].begin(), b[a[x]].end(), x))++;
                (*lower_bound(b[a[x + 1]].begin(), b[a[x + 1]].end(), x + 1))--;
                swap(a[x], a[x + 1]);
            }
        }
    }
    return 0;
}