[ARC065E]Manhattan Compass

2020-02-28
Atcoder

题意

在平面上有$n$个点,一开始指向$(s1,s2)$

若当前指向$(a,b)$,ab之间的曼哈顿距离和ac之间的曼哈顿距离相等,则可以指向$(a,c)$

数对均为无序,问能指向的数对个数,$n\leq 10^5$

题解

思路很简单,只要bfs即可,难点在如何快速确定下一个点的位置

不妨把平面旋转45度,$(x,y)$->$(x-y,x+y)$,曼哈顿距离转化为切比雪夫距离

这时可以把坐标离散化,对行和列开set

可以发现$d$一定,即一个点只要被处理一次即可,所以每次便利后在set中erase掉

这样还是不对,因为统计答案的时候前面的点可能还有贡献,那么我们再用vector统计答案

注意如果点在角上可能会计重($d_x=d_y$)

调试记录

在set里面二分要用set.lower_bound,切忌用lower_bound(set.begin(),set.end())

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#include <cstdio>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#define LL long long
const int maxn = 1e5 + 5;
using namespace std;
vector <pair<int, int> > X[maxn], Y[maxn];
set <pair<int, int> > sx[maxn], sy[maxn];
int n, s1, s2, d;
int x[maxn], y[maxn], lx[maxn], ly[maxn], ox[maxn], oy[maxn];
LL ans = 0;
queue <int> q;
bool vis[maxn];
void addx(int l, int f){
set <pair<int, int> >::iterator it = sx[l].lower_bound(make_pair(oy[f] - d, 0));
ans += 1ll * (upper_bound(X[l].begin(), X[l].end(), make_pair(oy[f] + d, n)) - lower_bound(X[l].begin(), X[l].end(), make_pair(oy[f] - d, 0)));
if (it == sx[l].end()) return;
while (oy[it->second] <= oy[f] + d && it != sx[l].end()){
if (!vis[it->second]) q.push(it->second), vis[it->second] = 1;
set <pair<int, int> >::iterator temp = it; ++temp;
sx[l].erase(it); it = temp;
}
}
void addy(int c, int f){
set <pair<int, int> >::iterator it = sy[c].lower_bound(make_pair(ox[f] - d, 0));
ans += 1ll * (upper_bound(Y[c].begin(), Y[c].end(), make_pair(ox[f] + d, n)) - lower_bound(Y[c].begin(), Y[c].end(), make_pair(ox[f] - d, 0)));
if (it == sy[c].end()) return;
while (ox[it->second] <= ox[f] + d && it != sy[c].end()){
if (!vis[it->second]) q.push(it->second), vis[it->second] = 1;
set <pair<int, int> >::iterator temp = it; ++temp;
sy[c].erase(it); it = temp;
}
}
map <pair<int, int>, int> m;
void bfs(){
q.push(s1); vis[s1] = 1;
while (!q.empty()){
int cur = q.front(); q.pop();
int l1 = lower_bound(lx + 1, lx + n + 1, ox[cur] - d) - lx;
int l2 = lower_bound(lx + 1, lx + n + 1, ox[cur] + d) - lx;
int c1 = lower_bound(ly + 1, ly + n + 1, oy[cur] - d) - ly;
int c2 = lower_bound(ly + 1, ly + n + 1, oy[cur] + d) - ly;
bool f1 = (lx[l1] == ox[cur] - d), f2 = (lx[l2] == ox[cur] + d);
bool f3 = (ly[c1] == oy[cur] - d), f4 = (ly[c2] == oy[cur] + d);
ans -= m[make_pair(l1, c1)] * f1 * f3;
ans -= m[make_pair(l2, c1)] * f2 * f3;
ans -= m[make_pair(l1, c2)] * f1 * f4;
ans -= m[make_pair(l2, c2)] * f2 * f4;
if (f1) addx(l1, cur);
if (f2) addx(l2, cur);
if (f3) addy(c1, cur);
if (f4) addy(c2, cur);
}
}
int _abs(int x){
return x < 0 ? -x : x;
}
int main(){
scanf("%d%d%d", &n, &s1, &s2);
for (int a, b, i = 1; i <= n; i++){
scanf("%d%d", &a, &b);
x[i] = a - b; y[i] = a + b;
ox[i] = x[i], oy[i] = y[i];
lx[i] = x[i], ly[i] = y[i];
}
sort(lx + 1, lx + n + 1); sort(ly + 1, ly + n + 1);
for (int i = 1; i <= n; i++){
x[i] = lower_bound(lx + 1, lx + n + 1, x[i]) - lx;
y[i] = lower_bound(ly + 1, ly + n + 1, y[i]) - ly;
X[x[i]].push_back(make_pair(oy[i], i));
Y[y[i]].push_back(make_pair(ox[i], i));
sx[x[i]].insert(make_pair(oy[i], i));
sy[y[i]].insert(make_pair(ox[i], i));
m[make_pair(x[i], y[i])]++;
}
for (int i = 1; i <= n; i++) sort(X[i].begin(), X[i].end());
for (int i = 1; i <= n; i++) sort(Y[i].begin(), Y[i].end());
d = max(_abs(ox[s1] - ox[s2]), _abs(oy[s1] - oy[s2]));
bfs();
printf("%lld\n", ans / 2ll);
return 0;
}