洛谷P2617 Dynamic Rankings

题意：带修区间第k大。

具体来说，就是带修主席树模板题。

解：唔...不离散化的话疯狂卡常都过不去，离散化就A了...虽然还是很慢。

把主席树换成树状数组套动态开点线段树即可。

#include <algorithm>
#include <cstdio>

const int N = 100010, M = 45000010;

struct Node {
    int f, x, y, z;
}node[N];

int a[N], now[N], rt[N], tot, n, p[N], tp, X[N << 1], temp;
int sum[M], ls[M], rs[M];
char str[3];
bool vis[N];

inline void read(int &x) {
    x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + c - 48;
        c = getchar();
    }
    return;
}

void add(int p, int v, int l, int r, int &o) {
    if(!o) {
        o = ++tot;
    }
    if(l == r) {
        sum[o] += v;
        return;
    }
    int mid = (l + r) >> 1;
    if(p <= mid) {
        add(p, v, l, mid, ls[o]);
    }
    else {
        add(p, v, mid + 1, r, rs[o]);
    }
    sum[o] = sum[ls[o]] + sum[rs[o]];
    return;
}

inline void insert(int p, int v, int id) {
    for(register int i = id; i <= n; i += i & (-i)) {
        add(p, v, 1, temp, rt[i]);
    }
    return;
}

inline int ask(int k, int L, int R) {
    for(register int i = L - 1; i >= 1; i -= i & (-i)) {
        vis[i] = 1;
        now[i] = rt[i];
        p[++tp] = i;
    }
    for(register int i = R; i >= 1; i -= i & (-i)) {
        if(!vis[i]) {
            vis[i] = 1;
            now[i] = rt[i];
            p[++tp] = i;
        }
    }
    //printf("ask : %d %d  k = %d 
", L, R, k);
    int l = 1, r = temp;
    while(l < r) {
        int mid = (l + r) >> 1, s = 0;
        for(register int i = L - 1; i >= 1; i -= i & (-i)) {
            s -= *(sum + (*(ls + (*(now + i)))));
            //printf("s -= %d 
", sum[ls[now[i]]]);
        }
        for(register int i = R; i >= 1; i -= i & (-i)) {
            s += *(sum + (*(ls + (*(now + i)))));
            //printf("s += %d 
", sum[ls[now[i]]]);
        }
        //printf("%d %d s = %d k = %d 
", l, r, s, k);
        if(k <= s) {
            for(register int i = 1; i <= tp; i++) {
                *(now + (*(p + i))) = *(ls + (*(now + (*(p + i)))));
            }
            r = mid;
        }
        else {
            for(register int i = 1; i <= tp; i++) {
                *(now + (*(p + i))) = *(rs + (*(now + (*(p + i)))));
            }
            k -= s;
            l = mid + 1;
        }
    }
    for(register int i = 1; i <= tp; i++) {
        vis[p[i]] = 0;
    }
    tp = 0;
    return r;
}

int main() {
    int m;
    read(n);
    read(m);
    for(register int i = 1; i <= n; i++) {
        read(a[i]);
        X[++temp] = a[i];
    }
    for(register int i = 1; i <= m; i++) {
        scanf("%s", str);
        read(node[i].x);
        read(node[i].y);
        if(str[0] == 'Q') {
            read(node[i].z);
        }
        else {
            node[i].f = 1;
            X[++temp] = node[i].y;
        }
    }
    std::sort(X + 1, X + temp + 1);
    temp = std::unique(X + 1, X + temp + 1) - X - 1;

    /*for(int i = 1; i <= temp; i++) {
        printf("%d ", X[i]);
    }
    puts("");*/

    for(int i = 1; i <= n; i++) {
        a[i] = std::lower_bound(X + 1, X + temp + 1, a[i]) - X;
        insert(a[i], 1, i);
        //printf("insert : %d %d 
", a[i], i);
    }
    for(int i = 1; i <= m; i++) {
        if(node[i].f) {
            node[i].y = std::lower_bound(X + 1, X + temp + 1, node[i].y) - X;
        }
    }
    for(register int i = 1, x, y, z; i <= m; i++) {
        x = node[i].x;
        y = node[i].y;
        if(!node[i].f) {
            z = node[i].z;
            printf("%d
", X[ask(z, x, y)]);
        }
        else {
            insert(a[x], -1, x);
            insert(y, 1, x);
            a[x] = y;
        }
    }
    return 0;
}