BZOJ4170 极光（CDQ分治 或 树套树）

传送门
BZOJ上的题目没有题面……

【样例输入】
3 5
2 4 3
Query 2 2
Modify 1 3
Query 2 2
Modify 1 2
Query 1 1
【样例输出】
2
3
3

这道题稍微分析一下就知道是求一个一个点曼哈顿距离小于k的的范围内的点的个数（把下标看做x，把值看做y）。然后我们只需要旋转一下坐标轴就变成了和“Mokia”或“简单题”一样的CDQ分治裸题了，求二维空间前缀和。
首先将询问按x排序，然后开始分治过程，计算左半区间对右半区间的贡献，然后就写完了呀！

然后是树套树做法，这个就更Naive了呀，直接套，反正就是求区间前缀和，乱搞搞就A了（不知道为什么对内存的需求不对啊……）。

CDQ版：

#include <cstdio>#include <algorithm>#include <iostream>#include <assert.h>#define LL int#define MAXM 300005#define MAXN 100005using namespace std;int a[MAXN];LL ans[MAXN];inline int Min(int a, int b) { return a < b ? a : b; }struct Querys{    int t, x, y, v, k; Querys(){}    Querys(int a, int b, int c, int d, int e):t(a),x(b),y(c),v(d),k(e){}    inline bool operator < (const Querys &r) const {        if(x == r.x && y == r.y) return t < r.t;        else if(x == r.x) return y < r.y;        else return x < r.x;    }} q[MAXM], nq[MAXM];int n, m, cnt = 0, qn, N;inline void GET(int &n) {    char c; n = 0;    do c=getchar(); while('0' > c || c > '9');    do n=n*10+c-'0',c=getchar(); while('0' <= c && c <= '9');}namespace BIT {    LL t[MAXM];    inline void Add(int x, int v) {        for(; x <= N; x += x&-x) t[x] += v;    }    inline LL Q(int x, LL sum = 0) {        for(; x; x -= x&-x) sum += t[x];        return sum;    }}void cdq(int L, int R) {    if(L >= R) return;    using namespace BIT;    int mid = (L + R) >> 1, lp = L, rp = mid+1;    for(int i = L; i <= R; ++ i)        if(q[i].t <= mid && q[i].k == 2) Add(q[i].y, 1);        else if(q[i].t > mid && q[i].k != 2) ans[q[i].v] += Q(q[i].y) * q[i].k;    for(int i = L; i <= R; ++ i) {        if(q[i].t <= mid && q[i].k == 2) Add(q[i].y, -1);        if(q[i].t <= mid) nq[lp ++] = q[i];        else nq[rp ++] = q[i];    }    for(int i = L; i <= R; ++ i) q[i] = nq[i];    cdq(L, mid); cdq(mid+1, R);}int main() {    GET(n); GET(m); N = 300000;    for(int i = 1; i <= n; ++ i) {        GET(a[i]);        q[++ cnt] = Querys(cnt, i + a[i], a[i] - i + 160000, 0, 2);    }    char op[10]; int u, v;    for(int i = 1; i <= m; ++ i) {        scanf("%s", op);        GET(u); GET(v);        if('M' == op[0]) {            a[u] = v;            q[++ cnt] = Querys(cnt, u + v, v - u + 160000, 0, 2);        }        else {            ++ qn; assert(a[u]-u-v+160000-1 > 0);            if(u+a[u]-v-1 > 0) q[++ cnt] = Querys(cnt, (u+a[u]-v-1), Min(a[u]-u+v+160000, N), qn, -1);            if(u+a[u]-v-1 > 0) q[++ cnt] = Querys(cnt, (u+a[u]-v-1), Min(a[u]-u-v+160000-1,N), qn, 1);            q[++ cnt] = Querys(cnt, (u+a[u]+v), Min(a[u]-u+v+160000,N), qn, 1);            q[++ cnt] = Querys(cnt, (u+a[u]+v), Min(a[u]-u-v+160000-1,N), qn, -1);        }    }    sort(q+1, q+cnt+1);    cdq(1, cnt);    for(int i = 1; i <= qn; ++ i)        printf("%d\n", ans[i]);    return 0;}

树套树版：

#include <cstdio>#define MAXN 1000005inline void GET(int &n) {    char c, f = 1; n = 0;    do if((c=getchar()) == '-') f = -1; while('0' > c || c > '9');    do n=n*10+c-'0',c=getchar(); while('0' <= c && c <= '9');    n *= f;}unsigned fix[MAXN * 20], sd = 2333;int key[MAXN * 20], s[MAXN * 20][2], sz[MAXN * 20], rt[MAXN], cnt, n, m, N, a[MAXN], num[MAXN];inline unsigned ran() { return sd = sd * sd + sd^27873; }inline void pushup(int x) {    sz[x] = sz[s[x][0]] + sz[s[x][1]] + num[x];}inline void rot(int &x, bool f) {    int t = s[x][f]; s[x][f] = s[t][f^1]; s[t][f^1] = x;    pushup(x); pushup(t); x = t;}void Insert(int &x, int v) {    if(!x) { x = ++ cnt; fix[x] = ran(); num[x] = sz[x] = 1; key[x] = v; return; }    if(v == key[x]) { ++ num[x]; pushup(x); return; }    bool f = key[x] < v;    Insert(s[x][f], v); pushup(x);    if(fix[s[x][f]] > fix[x]) rot(x, f);} int u, v, i;int Rank(int x, int v) {    if(!x) return 0;    if(key[x] < v) return sz[x] - sz[s[x][1]] + Rank(s[x][1], v);    else if(key[x] > v) return Rank(s[x][0], v);    else return sz[x] - sz[s[x][1]];}void Ins(int x, int v) {    for(; x <= N; x += x&-x)        Insert(rt[x], v);}int Q(int x, int y) {    int sum = 0;    for(; x > 0; x -= x&-x)        sum += Rank(rt[x], y);    return sum;}int main() {    GET(n); GET(m); N = 2*n + 100000;    for(int i = 1; i <= n; ++ i) {        GET(a[i]); Ins(i+a[i], a[i]-i);    }    char op[10];    for(i = 1; i <= m; ++ i) {        scanf("%s", op);        GET(u); GET(v);        if('M' == op[0]) {            a[u] = v; Ins(u+a[u], a[u]-u);        }        else {            int ans = Q((u+a[u]+v), a[u]-u+v) - Q((u+a[u]-v-1), a[u]-u+v) - Q((u+a[u]+v), a[u]-u-v-1) + Q((u+a[u]-v-1), a[u]-u-v-1);            printf("%d\n", ans);        }    }    return 0;}