CDQ分治总结

Posted 2022-09-03 buringstraw

目录

• 例题
  • 园丁的烦恼
  • 树状数组1
  • 陌上花开（三维偏序）
  • 摩基亚

同步：https://buringstraw.win/index.php/archives/50/

经过了一周的划水，我终于搞懂了cdq分治。

总的来说，cdq分治处理偏序问题就是

• 先把左边和右边当成一个完整的问题处理
• 然后把左边对右边的影响合并到右边

例题

园丁的烦恼

传送门

求静态区域内的点数，二维偏序模板题。

#include<cstdio> 
#include<algorithm>

const int MAXN = 500000 * 5 + 5;

//x,y:横纵坐标
//type:操作类型
//add:求矩形区域面积用几个矩形加加减减，所以add表示一下正负
//id:询问的id，因为一个询问拆成了好几个
//ans:存当前询问对应的答案
struct node 
    int x, y, type, add, id, ans;
 a[MAXN], tmp[MAXN];

int n, m, newp;
int ans[MAXN];

void add (int x, int y, int type, int add, int id, int ans) 
    a[++newp] = (node)x, y, type, add, id, ans;


bool cmp1 (node x, node y) 
    if (x.x == y.x) 
        if (x.y == y.y) //查到比自己y大的就停止
            return x.type < y.type;//修改在询问前
        
        return x.y < y.y;
    
    return x.x < y.x;


void cdq (int l, int r) 
    if (l == r) 
        return;
    
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    //左半边处理好了，把左半边所有修改加到右半边的询问中去
    int newl = l, newr = mid + 1, pos = l, ans = 0;
    //对于这个区间的上层区间，左边的x肯定都小于右边，所以这里按y排序后塞回a
    while (newl <= mid && newr <= r) //不能越界
        if (a[newl].y <= a[newr].y) //确保newl的操作在newr的范围内
            if (a[newl].type == 1) 
                ++ans;//是点，累加答案
            
            tmp[pos++] = a[newl++];
        
        else 
            if (a[newr].type == 2) 
                a[newr].ans += ans;//是询问，把前面统计的点加上
            
            tmp[pos++] = a[newr++];
        
    
    //没处理完的别剩着
    while (newl <= mid) 
        tmp[pos++] = a[newl++];
    
    while (newr <= r) 
        if (a[newr].type == 2) 
            a[newr].ans += ans;
        
        tmp[pos++] = a[newr++];
    
    //按y排好的结果装入a
    for (int i = l; i <= r; ++i) 
        a[i] = tmp[i];
    


int main (void) 
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) 
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y, 1, 0, 0, 0);
    
    for (int i = 1; i <= m; ++i) 
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        add(x2, y2, 2, 1, i, 0);
        add(x1 - 1, y2, 2, -1, i, 0);
        add(x2, y1 - 1, 2, -1, i, 0);
        add(x1 - 1, y1 - 1, 2, 1, i, 0);
    
    
    std::sort(a + 1, a + 1 + newp, cmp1);
    cdq(1, newp);
    
    for (int i = 1; i <= newp; ++i) 
        if (a[i].type == 2) 
            ans[a[i].id] += a[i].add * a[i].ans;
        
    
    
    for (int i = 1; i <= m; ++i) 
        printf("%d\n", ans[i]);
    
    return 0;

树状数组1

传送门

把操作出现的时间看做是第一维即可。

对于后面的数修改不会影响到前面的前缀和。

#include <cstdio> 

const int MAXN = 500000 + 5;

struct node 
    int x, y, id, type;
    friend bool operator <(node x, node y) 
        return x.x == y.x ? x.type < y.type :x.x < y.x;
    
 a[MAXN * 3], tmp[MAXN * 3];

int n, m, newp;
int ans[MAXN * 2];

void cdq(int l, int r) 
    if (l == r) 
        return;
    
    
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    
    int i = l, j = mid + 1, p = l, sum = 0;
    while (i <= mid && j <= r)   
        if (a[i] < a[j]) 
            if (a[i].type == 1) sum += a[i].y;
            tmp[p++] = a[i++];
        
        else 
            if (a[j].type == 2) 
                ans[a[j].id] += sum;
            
            tmp[p++] = a[j++];
        
    
    while (i <= mid) 
        tmp[p++] = a[i++];
    
    while (j <= r) 
        if (a[j].type == 2) 
            ans[a[j].id] += sum;
        
        tmp[p++] = a[j++];
    
    for (int i = l; i <= r; ++i) 
        a[i] = tmp[i];
    


int main (void) 
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) 
        scanf("%d", &a[++newp].y);
        a[newp].x = i;
        a[newp].type = 1;
    
    int opt, x, y, z, anscnt = 0;
    for (int i = 1; i <= m; ++i)  
        scanf("%d", &opt);
        if (opt == 1) 
            scanf("%d%d", &x, &y);
            a[++newp].x = x;
            a[newp].y = y;
            a[newp].type = 1;
        
        else 
            ++anscnt;
            scanf("%d%d", &x, &y);
            a[++newp].x = x - 1;
            a[newp].id = anscnt * 2 - 1;
            a[newp].type = 2;
            a[++newp].x = y;
            a[newp].id = anscnt * 2;
            a[newp].type = 2;
        
    
    cdq(1, newp);
    for (int i = 1; i <= anscnt; ++i) 
        printf("%d\n", ans[i * 2] - ans[i * 2 - 1]);
    
    return 0;

陌上花开（三维偏序）

为什么我要把“陌上花开”写在前面呢？因为这样看起来比较帅

第一维排序，第二位像原来那样比大小，然后搞个值域树状数组来统计每个0~x里的第三维

要去重

#include <cstdio>
#include <algorithm>

const int MAXN = 200000 + 5;
namespace sz 
    int n;
    int lowbit (int x)return x & (-x);
    int c[MAXN * 2];
    void add (int x, int k) 
        while (x <= n) 
            c[x] += k;
            x += lowbit(x);
        
    
    int query (int x) 
        int ans = 0;
        while (x > 0) 
            ans += c[x];
            x -= lowbit(x);
        
        return ans;
    



struct node 
    int a, b, c, id;
 a[MAXN], tmp[MAXN];

int n, k, newp;
int size[MAXN], ans[MAXN], num[MAXN];

bool cmp1 (node x, node y) 
    return x.a == y.a ? (x.b == y.b ? x.c < y .c : x.b < y.b) : x.a < y.a;


void cdq(int l, int r) 
    if (l == r) return;
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    int i = l, j = mid + 1, p = l;
    while (i <= mid && j <= r) 
        if (a[i].b <= a[j].b) 
            sz::add(a[i].c, size[a[i].id]);
            tmp[p++] = a[i++];
        
        else 
            ans[a[j].id] += sz::query(a[j].c);
            tmp[p++] = a[j++];
        
    
    while (j <= r) 
        ans[a[j].id] += sz::query(a[j].c);
        tmp[p++] = a[j++];
    
    for (int h = l; h < i; ++h) 
        sz::add(a[h].c, -size[a[h].id]);
    
    while (i <= mid) 
        tmp[p++] = a[i++];
    
    for(int i = l; i <= r; ++i) 
        a[i] = tmp[i];
    


int main (void) 
    scanf("%d%d", &n, &k);
    sz::n = k;
    for (int i = 1; i <= n; ++i) 
        scanf("%d%d%d", &a[i].a, &a[i].b, &a[i].c);
    
    std::sort(a + 1, a + 1 + n, cmp1);
    for (int i = 1; i <= n; ++i) 
        if (a[i].a != a[i - 1].a || a[i].b != a[i - 1].b || a[i].c != a[i - 1].c) 
            tmp[++newp] = a[i];
        
        ++size[newp];
    
    for (int i =1; i <= newp; ++i) 
        a[i] = tmp[i];
        a[i].id = i;
    
    cdq(1, newp);
    for (int i = 1; i <=newp; ++i) 
        num[ans[a[i].id] + size[a[i].id] - 1] += size[a[i].id];//除了ans里的数量，还要加上完全相同的数量 
    
    for (int i = 0; i < n; ++i) 
        printf("%d\n", num[i]);
    
    return 0;

摩基亚

传送门

其实这是 Nokia 哒

本来是个二维，加上时间顺序就是三维了。

树状数组下标不能为0

#include <cstdio>
#include <algorithm>

const int MAXN = 1700000 + 5;
namespace sz 
    int n;
    int lowbit (int x)return x & (-x);
    int c[MAXN * 2];
    void add (int x, int k) 
        while (x <= n) 
            c[x] += k;
            x += lowbit(x);
        
    
    int query (int x) 
        int ans = 0;
        while (x > 0) 
            ans += c[x];
            x -= lowbit(x);
        
        return ans;
    
    void clear (int x) 
        while (x <= n) 
            c[x] = 0;
            x += lowbit(x);
               
    

struct node 
    int x, y, id, type, val;
    friend bool operator <(node x, node y) 
        return x.x == y.x ? x.y == y.y ? x.type < y.type : x.y < y.y : x.x < y.x;
    
 a[MAXN], tmp[MAXN];

int n, newp, newq;
int ans[MAXN];

void cdq(int l, int r) 
    if (l == r) 
        return;
    
    
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    int i = l, j = mid + 1, p = l;
    
    while (i <= mid && j <= r) 
        if (a[i].x <= a[j].x) 
            if(a[i].type == 1)sz::add(a[i].y, a[i].val);    
            tmp[p++] = a[i++];
        
        else 
            if (a[j].type == 2)ans[a[j].id] += sz::query(a[j].y);
            tmp[p++] = a[j++];
        
    
    while (i <= mid) 
        tmp[p++] = a[i++];
    
    while (j <= r) 
        if (a[j].type == 2) 
            ans[a[j].id] += sz::query(a[j].y);
        
        tmp[p++] = a[j++];
    
    for (int k = l; k <= mid; ++k) 
        if (a[k].type == 1) sz::clear(a[k].y);
    
    for (int i = l; i <= r; ++i) 
        a[i] = tmp[i];
    


void read (int &x) 
    x = 0;
    int k = 1;
    int t = getchar();
    while (t > '9' || t < '0') 
        if (t == '-') k = -1;
        t = getchar();
    
    while (t >= '0' && t <= '9') 
        x *= 10;
        x += (t - '0');
        t = getchar();
    
    x *= k;


int main (void) 
    read(n);read(n);
    ++n;
    sz::n = n;
    int opt;
    read(opt);
    while (opt != 3) 
        if (opt == 1) 
            ++newp;
            read(a[newp].x);read(a[newp].y);read(a[newp].val);
            ++a[newp].x;++a[newp].y;
            a[newp].type = 1;
        
        else 
            int x1, x2, y1, y2;
            read(x1);read(y1);read(x2);read(y2);
            ++x1;++x2;++y1;++y2;
            a[++newp].x = x2;a[newp].y = y2;a[newp].type = 2;a[newp].id = ++newq;
            a[++newp].x = x1 -1;a[newp].y = y2;a[newp].type = 2;a[newp].id = ++newq;
            a[++newp].x = x2;a[newp].y = y1 - 1;a[newp].type = 2;a[newp].id = ++newq;
            a[++newp].x = x1 - 1;a[newp].y = y1 - 1;a[newp].type = 2;a[newp].id = ++newq;
        
        read(opt);
    
    //std::sort(a + 1, a + 1 + newp, cmp1);
    cdq(1, newp);
    for (int i = 1; i <= newq; i += 4) 
        printf("%d\n", ans[i] - ans[i + 1] - ans[i + 2] + ans[i + 3]);
    
    return 0;