[Violet]天使玩偶/SJY摆棋子
解题思路
用CDQ分治开了氧气跑过。
将输入给的顺序作为第一维的时间,x为第二维,y为第三维。对于距离一个询问(ax,ay),将询问分为四块,左上,右上,左下,右下,对于坐下,左下的dist即为ax+ay-max(bx+by)。所以只要查询时间小于自己的点里x+y最大的即可。对于其他四块,都可以转为左下,各自跑一遍CDQ。
代码如下
#include <bits/stdc++.h>
using namespace std;
int n, m;
const int N = 1e6+5;
struct T{
int t, x, y;
bool f;
T(){}
T(int t, int x, int y, bool f): t(t), x(x), y(y), f(f){}
}v[N], t[N], a[N];
int c[N];
int lowbit(int x)
{
return x & (-x);
}
void update(int x, int v)
{
for(int i = x; i < N; i += lowbit(i))
c[i] = max(c[i], v);
}
int query(int x)
{
int ans = -1;
for(int i = x; i > 0; i -= lowbit(i)){
ans = max(c[i], ans);
}
return ans;
}
void clear(int x)
{
for(int i = x; i < N; i += lowbit(i))
c[i] = 0;
}
int ans[N];
void CDQ(int l, int r)
{
if(l == r)
return;
int mid = (l + r) / 2;
CDQ(l, mid);
CDQ(mid + 1, r);
for(int i = l; i <= r; i ++)
t[i] = v[i];
int x = l, y = mid + 1;
int cnt = l - 1;
while(x <= mid || y <= r){
if(x <= mid && y <= r){
if(t[x].x <= t[y].x){
if(!t[x].f)update(t[x].y, t[x].x + t[x].y);
v[++cnt] = t[x];
++x;
}
else {
int r = query(t[y].y);
if(t[y].f && r) ans[t[y].t] = min(ans[t[y].t], t[y].x + t[y].y - r);
v[++cnt] = t[y];
++y;
}
}
else if(x <= mid){
if(!t[x].f)update(t[x].y, t[x].x + t[x].y);
v[++cnt] = t[x];
++x;
}
else {
int r = query(t[y].y);
if(t[y].f && r) ans[t[y].t] = min(ans[t[y].t], t[y].x + t[y].y - r);
v[++cnt] = t[y];
++y;
}
}
for(int i = l; i <= mid; i ++)
if(!t[i].f)clear(t[i].y);
}
int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i ++){
int x, y;
scanf("%d%d", &x, &y);
v[i] = T(0, x + 1, y + 1, 0);
}
for(int i = 1; i <= m; i ++){
int t, x, y;
scanf("%d%d%d", &t, &x, &y);
v[n+i] = T(i, x + 1, y + 1, t - 1);
}
int lx, ly;
lx = ly = 0;
for(int i = 1; i <= n + m; i ++){
a[i] = v[i];
lx = max(lx, v[i].x + 1);
ly = max(ly, v[i].y + 1);
}
memset(ans, 0x3f, sizeof(ans));
CDQ(1, n + m);
for(int i = 1; i <= n + m; i ++){
v[i] = a[i];
v[i].x = lx - v[i].x;
}
CDQ(1, n + m);
for(int i = 1; i <= n + m; i ++){
v[i] = a[i];
v[i].y = ly - v[i].y;
}
CDQ(1, n + m);
for(int i = 1; i <= n + m; i ++){
v[i] = a[i];
v[i].x = lx - v[i].x;
v[i].y = ly - v[i].y;
}
CDQ(1, n + m);
for(int i = 1; i <= m; i ++){
if(a[i + n].f)
printf("%d\n", ans[i]);
}
return 0;
}