# 题目大意

1.在平面上加入一条线段。记第i条被插入的线段的标号为i。
2.给定一个数k,询问与直线 x = k相交的线段中，交点最靠上的线段的编号。

# 代码

#include <algorithm>
#include <cctype>
#include <cstdio>
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-')
f = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int N, M;
struct line {
double k, b;
int id;
line(int x0 = 0, int y0 = 0, int x1 = 0, int y1 = 0, int ID = 0) {
id = ID;
if (x0 == x1)
k = 0, b = std::max(y0, y1);
else
k = (double)(y0 - y1) / (x0 - x1), b = (double)y0 - k * x0;
}
double getf(int x) { return k * x + b; };
};
bool cmp(line a, line b, int x) {
if (!a.id)
return 1;
return a.getf(x) != b.getf(x) ? a.getf(x) < b.getf(x) : a.id < b.id;
}
const int maxn = 50010;
line t[maxn << 2];
line query(int k, int l, int r, int x) {
if (l == r)
return t[k];
int mid = (l + r) >> 1;
line tmp;
if (x <= mid)
tmp = query(k << 1, l, mid, x);
else
tmp = query(k << 1 | 1, mid + 1, r, x);
return cmp(t[k], tmp, x) ? tmp : t[k];
}
void insert(int k, int l, int r, line x) {
if (!t[k].id)
t[k] = x;
if (cmp(t[k], x, l))
std::swap(t[k], x);
if (l == r || t[k].k == x.k)
return;
int mid = (l + r) >> 1;
double X = (t[k].b - x.b) / (x.k - t[k].k);
if (X < l || X > r)
return;
if (X <= mid)
insert(k << 1, l, mid, t[k]), t[k] = x;
else
insert(k << 1 | 1, mid + 1, r, x);
}
void Insert(int k, int l, int r, int x, int y, line v) {
if (x <= l && r <= y) {
insert(k, l, r, v);
return;
}
int mid = (l + r) >> 1;
if (x <= mid)
Insert(k << 1, l, mid, x, y, v);
if (y > mid)
Insert(k << 1 | 1, mid + 1, r, x, y, v);
}
#define p1 39989
#define p2 1000000000
int main() {
#ifdef D
freopen("input", "r", stdin);
#endif
N = 50000;
int las, cnt = 0;
while (M--) {
if (opt == 0) {
x = (x + las - 1) % p1 + 1;
las = query(1, 1, N, x).id;
printf("%d\n", las);
}
if (opt == 1) {
x0 = (x0 + las - 1) % p1 + 1;
y0 = (y0 + las - 1) % p2 + 1;
x1 = (x1 + las - 1) % p1 + 1;
y1 = (y1 + las - 1) % p2 + 1;
if (x0 > x1)
std::swap(x0, x1), std::swap(y0, y1);
Insert(1, 1, N, x0, x1, line(x0, y0, x1, y1, ++cnt));
}
}
return 0;
}


posted on 2017-03-02 08:14  蒟蒻konjac  阅读(351)  评论(0编辑  收藏

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