【BZOJ 1901】【ZJU 2112】Dynamic Rankings

http://www.lydsy.com/JudgeOnline/problem.php?id=1901
重新用整体二分写了一下。
整体二分的思想详见论文。
貌似带修区间k大和静态区间k大都是\(O(n\log^2n)\)
cdq是对时间分治,整体二分是对答案分治。
对于动态区间k大怎么实现我还想了很长时间呢,感觉模板题都想这么长时间。。。
动态区间k大的关键步骤是把修改变成两个操作,删除和插入。
这样在一个分治函数内部就可以扫时间轴然后在特定权值范围内的删除和插入分别在bits上-1和+1。
相比较在线bits套主席树,离线整体二分跑得飞快。
离散化后跑得能快一点,不离散的话时间复杂度就是\(O(n\log n\log 10^9)\),离散的话时间复杂度是\(O(n\log^2n)\)

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

const int N = 10003;

struct node {int x, y, k, op, id;} B[N << 2], q1[N << 2], q2[N << 2];
int tot = 0, a[N], n, m, bits[N << 1], top = 0, H[N << 1], cnt = 0;
int ans[N], id[N << 2], anstot = 0, que[N << 1], q1len, q2len;

void add(int x, int flag) {for (; x <= n; x += (x & (-x))) bits[x] += flag;}
int sum(int x) {int ret = 0; for (; x; x -= (x & (-x))) ret += bits[x]; return ret;}

void solve(int p, int q, int l, int r) {
    if (p > q) return;
    if (l == r) {
        for (int i = p; i <= q; ++i)
            if (B[i].op == 3)
                ans[B[i].id] = l;
        return;
    }
    
    int mid = (l + r) >> 1;
    for (int i = p; i <= q; ++i) {
        if (B[i].op != 3) {
            if (B[i].y <= mid)
                add(B[i].x, B[i].op);
        }
        else que[i] = sum(B[i].y) - sum(B[i].x - 1);
    }
    
    for (int i = p; i <= q; ++i)
        if (B[i].op != 3 && B[i].y <= mid) add(B[i].x, -B[i].op);
    
    q1len = q2len = 0;
    for (int i = p; i <= q; ++i)
        if (B[i].op != 3) {
            if (B[i].y > mid) q2[++q2len] = B[i];
            else q1[++q1len] = B[i];
        } else {
            if (que[i] >= B[i].k) q1[++q1len] = B[i];
            else B[i].k -= que[i], q2[++q2len] = B[i];
        }
    
    int tmp = p - 1;
    for (int i = 1; i <= q1len; ++i)
        B[++tmp] = q1[i];
    for (int i = 1; i <= q2len; ++i)
        B[++tmp] = q2[i];
    
    tmp = p + q1len - 1;
    solve(p, tmp, l, mid);
    solve(tmp + 1, q, mid + 1, r);
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        B[++tot] = (node) {i, a[i], 0, 1, 0};
        H[++cnt] = a[i];
    }
    
    int x, y, k;
    char c[2];
    for (int i = 1; i <= m; ++i) {
        scanf("%s", c);
        if (c[0] == 'Q') {
            scanf("%d%d%d", &x, &y, &k);
            B[++tot] = (node) {x, y, k, 3, ++anstot};
        } else {
            scanf("%d%d", &x, &y);
            B[++tot] = (node) {x, a[x], 0, -1, 0};
            B[++tot] = (node) {x, y, 0, 1, 0};
            H[++cnt] = (a[x] = y);
        }
    }
    
    stable_sort(H + 1, H + cnt + 1);
    cnt = unique(H + 1, H + cnt + 1) - H;
    for (int i = 1; i <= tot; ++i)
        if (B[i].op != 3)
            B[i].y = lower_bound(H + 1, H + cnt, B[i].y) - H;
    
    solve(1, tot, 1, cnt - 1);
    
    for (int i = 1; i <= anstot; ++i) printf("%d\n", H[ans[i]]);
    return 0;
}
posted @ 2017-02-01 12:33  abclzr  阅读(...)  评论(...编辑  收藏