标签：++ 表示 完整 排序 val can print list ring
目录
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
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;
}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;
}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;
}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;
}标签：++ 表示 完整 排序 val can print list ring
原文地址：https://www.cnblogs.com/buringstraw/p/11420927.html
