SCUT - 484 - 平面上的点 - 数据结构
一开始想的是按固定斜率的直线从无穷扫下来，但是一直都WA，不知道是哪里错了还是精度问题？
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
int dcmp(long double x, long double y) {
if(fabs(x - y) <= 1e-14)
return 0;
return x < y ? -1 : 1;
}
int n;
ll q;
int ch[MAXN][2];
int val[MAXN];
int dat[MAXN], siz[MAXN], cnt[MAXN];
int tot, root;
inline void Init() {
tot = 0;
root = 0;
}
inline int NewNode(int v) {
val[++tot] = v, dat[tot] = rand();
ch[tot][0] = ch[tot][1] = 0;
siz[tot] = 1, cnt[tot] = 1;
return tot;
}
inline void PushUp(int id) {
siz[id] = siz[ch[id][0]] + siz[ch[id][1]] + cnt[id];
}
inline void Rotate(int &id, int d) {
int temp = ch[id][d ^ 1];
ch[id][d ^ 1] = ch[temp][d];
ch[temp][d] = id;
id = temp;
PushUp(ch[id][d]), PushUp(id);
}
inline void Insert(int &id, int v) {
if(!id)
id = NewNode(v);
else {
if(v == val[id])
++cnt[id];
else {
int d = v < val[id] ? 0 : 1;
Insert(ch[id][d], v);
if(dat[id] < dat[ch[id][d]])
Rotate(id, d ^ 1);
}
PushUp(id);
}
}
//找严格小于v的点的个数
int GetRank(int id, int v) {
if(!id)
return 0;
else {
if(v == val[id])
return siz[ch[id][0]];
else if(v < val[id])
return GetRank(ch[id][0], v);
else
return siz[ch[id][0]] + cnt[id] + GetRank(ch[id][1], v);
}
}
const ll INF = 1e9;
struct Point {
long double x, y;
Point() {}
Point(long double x, long double y): x(x), y(y) {}
Point Rotate(long double A) {
return Point(x * cos(A) - y * sin(A), x * sin(A) + y * cos(A));
}
bool operator<(const Point &p)const {
return (dcmp(x, p.x) != 0) ? (x < p.x) : (y < p.y);
}
} p[MAXN], p2[MAXN];
int nxt[MAXN];
bool check(ll k) {
Init();
long double A = atan2(1.0, 1.0 * k);
for(int i = 1; i <= n; ++i)
p2[i] = p[i].Rotate(A);
//按斜率排序
sort(p2 + 1, p2 + 1 + n);
for(int i = 1; i <= n; ++i) {
p2[i] = p2[i].Rotate(-A);
p2[i].x = round(p2[i].x);
p2[i].y = round(p2[i].y);
}
ll sum = 0;
for(int i = 1; i <= n; ++i) {
int res = GetRank(root, (int)p2[i].x);;
sum += res;
if(sum >= q)
return true;
Insert(root, (int)p2[i].x);
}
return false;
}
ll CASE() {
scanf("%d%lld", &n, &q);
long double maxy = -INF, miny = INF;
for(int i = 1; i <= n; ++i) {
cin>>p[i].x>>p[i].y;
//scanf("%lf%lf", &p[i].x, &p[i].y);
if(p[i].y > maxy)
maxy = p[i].y;
if(p[i].y < miny)
miny = p[i].y;
}
ll L = -round(maxy - miny), R = round(maxy - miny), M;
while(1) {
M = (L + R) >> 1;
if(L == M) {
if(check(L))
return L;
if(check(R))
return R;
return INF;
}
if(check(M))
R = M;
else
L = M + 1;
}
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int T;
while(~scanf("%d", &T)) {
while(T--) {
ll res = CASE();
if(res >= INF)
puts("INF");
else
printf("%lld\n", res);
}
}
return 0;
}
事实上枚举斜率之后对式子变形：
\(\frac{y_1-y_2}{x_1-x_2}<=k\)
不妨设x1>x2
\(y_1-y_2<=k(x_1-x_2)\)
即：
\(y_1-kx_1<=y_2-kx_2\)
即满足 \(x1>x2\) 且 \(y_1-kx_1<=y_2-kx_2\) 的数对的个数。lzf大佬说是逆序对，太强了。
平衡树卡过去非常勉强：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
int n;
ll q;
int ch[MAXN][2];
ll val[MAXN];
int dat[MAXN], siz[MAXN], cnt[MAXN];
int tot, root;
inline void Init() {
tot = 0;
root = 0;
}
inline int NewNode(ll v) {
val[++tot] = v, dat[tot] = rand();
ch[tot][0] = ch[tot][1] = 0;
siz[tot] = 1, cnt[tot] = 1;
return tot;
}
inline void PushUp(int id) {
siz[id] = siz[ch[id][0]] + siz[ch[id][1]] + cnt[id];
}
inline void Rotate(int &id, int d) {
int temp = ch[id][d ^ 1];
ch[id][d ^ 1] = ch[temp][d];
ch[temp][d] = id;
id = temp;
PushUp(ch[id][d]), PushUp(id);
}
inline void Insert(int &id, ll v) {
if(!id)
id = NewNode(v);
else {
if(v == val[id])
++cnt[id];
else {
int d = v < val[id] ? 0 : 1;
Insert(ch[id][d], v);
if(dat[id] < dat[ch[id][d]])
Rotate(id, d ^ 1);
}
PushUp(id);
}
}
int GetRank(int id, ll v) {
if(!id)
return 0;
else {
if(v == val[id])
return siz[ch[id][1]] + cnt[id];
else if(v < val[id])
return siz[ch[id][1]] + cnt[id] + GetRank(ch[id][0], v);
else
return GetRank(ch[id][1], v);
}
}
const int INF = 1e9;
struct Point {
int x, y;
ll z;
Point() {}
Point(int x, int y): x(x), y(y) {}
bool operator<(const Point &p)const {
return (x != p.x) ? (x < p.x) : (y < p.y);
}
} p[MAXN];
bool check(int k) {
//printf("k=%d\n", k);
Init();
for(int i = 1; i <= n; ++i) {
p[i].z = 1ll * p[i].y - 1ll * k * p[i].x;
//printf("p[%d].z=%lld%c", i, p[i].z, " \n"[i == n]);
//printf("p[%d].x=%d%c", i, p[i].x, " \n"[i == n]);
}
ll sum = 0;
for(int i = 1, nxt; i <= n; i = nxt) {
for(nxt = i + 1; nxt <= n && p[nxt].x == p[i].x; ++nxt);
for(int j = i; j < nxt; ++j) {
//int gr = GetRank(root, p[j].z);
//printf("j=%d ,gr=%d\n", j, gr);
sum += GetRank(root, p[j].z);
//printf("sum=%lld\n", sum);
// if(sum >= q)
// return true;
}
for(int j = i; j < nxt; ++j)
Insert(root, p[j].z);
}
if(sum >= q)
return true;
return false;
}
int solve() {
scanf("%d%lld", &n, &q);
int maxy = -INF, miny = INF;
for(int i = 1; i <= n; ++i) {
scanf("%d%d", &p[i].x, &p[i].y);
if(p[i].y > maxy)
maxy = p[i].y;
if(p[i].y < miny)
miny = p[i].y;
}
sort(p + 1, p + 1 + n);
/*for(int i = 1; i <= n; ++i) {
//p[i].z = 1ll * p[i].y - 1ll * k * p[i].x;
//printf("p[%d].z=%lld%c", i, p[i].z, " \n"[i == n]);
printf("p[%d].x=%d%c", i, p[i].x, " \n"[i == n]);
}*/
int L = -(maxy - miny), R = maxy - miny, M;
while(1) {
M = (L + R) >> 1;
if(L == M) {
if(check(L))
return L;
if(check(R))
return R;
return INF;
}
if(check(M))
R = M;
else
L = M + 1;
}
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int T;
while(~scanf("%d", &T)) {
while(T--) {
int res = solve();
if(res >= INF)
puts("INF");
else
printf("%d\n", res);
}
}
return 0;
}
求逆序对，用树状数组的话，是这样思考：把其中一维离散化，用树状数组来表示，按另一维顺序插入。
下面是把z离散化，按x从小到大插入，那么每次合法的就是前面的比它小的x2里面z2>=z1的，这样就先求出<=z1-1的个数，然后用已经插入的数的个数i-1减去，这里要注意相同的x的处理，600ms：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
int n;
ll q;
int bit[MAXN];
inline void Init() {
memset(bit, 0, sizeof(bit[0]) * (n + 1));
}
inline int Sum(int x) {
int res = 0;
while(x) {
res += bit[x];
x -= x & -x;
}
return res;
}
inline void Add(int x, int v) {
while(x <= n) {
bit[x] += v;
x += x & -x;
}
}
struct Point {
int x, y;
ll z;
Point() {}
Point(int x, int y): x(x), y(y) {}
bool operator<(const Point &p)const {
return x < p.x;
}
} p[MAXN];
ll pz[MAXN];
bool check(int k) {
Init();
for(int i = 1; i <= n; ++i)
pz[i] = p[i].z = 1ll * p[i].y - 1ll * k * p[i].x;
sort(pz + 1, pz + 1 + n);
int pzn = unique(pz + 1, pz + 1 + n) - (pz + 1);
for(int i = 1; i <= n; ++i)
p[i].z = lower_bound(pz + 1, pz + 1 + pzn, p[i].z) - pz;
ll sum = 0;
for(int i = 1, nxt; i <= n; i = nxt) {
for(nxt = i + 1; nxt <= n && p[nxt].x == p[i].x; ++nxt);
for(int j = i; j < nxt; ++j) {
sum += ((ll)i - 1) - Sum(p[j].z - 1);
if(sum >= q)
return true;
}
for(int j = i; j < nxt; ++j)
Add(p[j].z, 1);
}
return false;
}
const int INF = 1e9;
int solve() {
scanf("%d%lld", &n, &q);
int maxy = -INF, miny = INF;
for(int i = 1; i <= n; ++i) {
scanf("%d%d", &p[i].x, &p[i].y);
if(p[i].y > maxy)
maxy = p[i].y;
if(p[i].y < miny)
miny = p[i].y;
}
sort(p + 1, p + 1 + n);
int L = -(maxy - miny), R = maxy - miny, M;
while(1) {
M = (L + R) >> 1;
if(L == M) {
if(check(L))
return L;
if(check(R))
return R;
return INF;
}
if(check(M))
R = M;
else
L = M + 1;
}
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int T;
while(~scanf("%d", &T)) {
while(T--) {
int res = solve();
if(res >= INF)
puts("INF");
else
printf("%d\n", res);
}
}
return 0;
}
离散化x的话，会更快，300ms，毕竟不用每次都对z进行一次unique然后lowerbound，的确少了一半的常数。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
int n;
ll q;
int bit[MAXN];
inline void Init() {
memset(bit, 0, sizeof(bit[0]) * (n + 1));
}
inline int Sum(int x) {
int res = 0;
while(x) {
res += bit[x];
x -= x & -x;
}
return res;
}
inline void Add(int x) {
while(x <= n) {
++bit[x];
x += x & -x;
}
}
struct Point {
int x, y, idx;
ll z;
Point() {}
Point(int x, int y): x(x), y(y) {}
bool operator<(const Point &p)const {
return z == p.z ? idx<p.idx: z > p.z;
}
} p[MAXN];
int px[MAXN], pxn;
bool check(int k) {
Init();
for(int i = 1; i <= n; ++i)
p[i].z = p[i].y - 1ll * k * p[i].x;
sort(p + 1, p + 1 + n);
ll sum = 0;
for(int i = 1; i <= n; ++i) {
sum += Sum(p[i].idx - 1);
if(sum >= q)
return true;
Add(p[i].idx);
}
return false;
}
const int INF = 1e9;
int solve() {
scanf("%d%lld", &n, &q);
int maxy = -INF, miny = INF;
for(int i = 1; i <= n; ++i) {
scanf("%d%d", &p[i].x, &p[i].y);
px[i] = p[i].x;
if(p[i].y > maxy)
maxy = p[i].y;
if(p[i].y < miny)
miny = p[i].y;
}
sort(px + 1, px + 1 + n);
pxn = unique(px + 1, px + 1 + n) - (px + 1);
for(int i = 1; i <= n; ++i)
p[i].idx = lower_bound(px + 1, px + 1 + pxn, p[i].x) - px;
int L = miny - maxy, R = maxy - miny, M;
while(1) {
M = (L + R) >> 1;
if(L == M) {
if(check(L))
return L;
if(check(R))
return R;
return INF;
}
if(check(M))
R = M;
else
L = M + 1;
}
}
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
#endif // Yinku
int T;
while(~scanf("%d", &T)) {
while(T--) {
int res = solve();
if(res >= INF)
puts("INF");
else
printf("%d\n", res);
}
}
return 0;
}
