UVA 10869 - Brownie Points II(树阵)

UVA 10869 - Brownie Points II

题目链接

题意：平面上n个点，两个人，第一个人先选一条经过点的垂直x轴的线。然后还有一个人在这条线上穿过的点选一点作垂直该直线的线，然后划分出4个象限，第一个人得到分数为1。3象限，第二个人为二四象限。问第一个个人按最优取法，能得到最小分数的最大值，和这个值下还有一个人的得分可能情况

思路：树状数组，能够枚举一点，假设能求出右上和左下点的个数就好办了，其有用一个树状数组，把y坐标离散化掉，然后记录进来，然后把点按x从左往右，每次删掉点后查询当前高度向上的点的个数，这样就能得到右上角点的个数，同理也能求左下角，能处理这步就好办了。最后注意答案要取重而且递增，用set就ok了

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;

#define lowbit(x) (x&(-x))

const int N = 200005;

struct Point {
int x, y, rank, ans, Mi;
int xn, yn;
} p
;

int bit
;
set<int> out;

void add(int x, int v) {
while (x < N) {
bit[x] += v;
x += lowbit(x);
}
}

int get(int x) {
int ans = 0;
while (x > 0) {
ans += bit[x];
x -= lowbit(x);
}
return ans;
}

int get(int l, int r) {
return get(r) - get(l - 1);
}

bool cmpy(Point a, Point b) {
return a.y < b.y;
}

bool cmpx(Point a, Point b) {
if (a.x == b.x) return a.y > b.y;
return a.x < b.x;
}

bool cmpx2(Point a, Point b) {
if (a.x == b.x) return a.y < b.y;
return a.x > b.x;
}

int n, top;

void init() {
memset(bit, 0, sizeof(bit));
for (int i = 0; i < n; i++) {
scanf("%d%d", &p[i].x, &p[i].y);
p[i].ans = 0;
p[i].xn = p[i].yn = 0;
}
sort(p, p + n, cmpy);
}

void solve() {
top = 1;
for (int i = 0; i < n; i++) {
if (i && p[i].y != p[i - 1].y)
top++;
p[i].rank = top;
add(p[i].rank, 1);
}
for (int i = 0; i < n; i++) {
int j;
int len = 0;
for (j = i; p[i].y == p[j].y && j < n; j++)
len++;
for (j = i; p[i].y == p[j].y && j < n; j++)
p[j].yn = len;
i = j - 1;
}
sort(p, p + n, cmpx);
for (int i = 0; i < n; i++) {
add(p[i].rank, -1);
p[i].ans += get(p[i].rank + 1, top);
}
for (int i = 0; i < n; i++) {
int j;
int len = 0;
for (j = i; p[i].x == p[j].x && j < n; j++)
len++;
for (j = i; p[i].x == p[j].x && j < n; j++)
p[j].xn = len;
i = j - 1;
}
memset(bit, 0, sizeof(bit));
for (int i = 0; i < n; i++)
add(p[i].rank, 1);
sort(p, p + n, cmpx2);
for (int i = 0; i < n; i++) {
add(p[i].rank, -1);
p[i].ans += get(1, p[i].rank - 1);
}
for (int i = 0; i < n; i++) {
int j;
int Min = 1000000000;
for (j = i; p[i].x == p[j].x && j < n; j++)
Min = min(Min, p[j].ans);
for (j = i; p[i].x == p[j].x && j < n; j++)
p[j].Mi = Min;
i = j - 1;
}
int Max = 0;
out.clear();
for (int i = 0; i < n; i++) {
if (p[i].Mi != p[i].ans) continue;
if (p[i].ans > Max) {
Max = p[i].ans;
out.clear();
out.insert(n - p[i].ans - p[i].xn - p[i].yn + 1);
}
else if (p[i].ans == Max) {
out.insert(n - p[i].ans - p[i].xn - p[i].yn + 1);
}
}
printf("Stan: %d; Ollie:", Max);
for (set<int>::iterator it = out.begin(); it != out.end(); it++)
printf(" %d", *it);
printf(";\n");
}

int main() {
while (~scanf("%d", &n) && n) {
init();
solve();
}
return 0;
}