记录编号 |
196438 |
评测结果 |
AAAAAAAAAAAA |
题目名称 |
[USACO Jan07]考试 |
最终得分 |
100 |
用户昵称 |
ppfish |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.020 s |
提交时间 |
2015-10-21 19:34:12 |
内存使用 |
3.79 MiB |
显示代码纯文本
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int bufsize = 1 << 20;
char buf[bufsize], *ins = buf, *ous = buf;
inline char getc()
{
if (ins == ous) ins = buf, ous = buf + fread(buf, 1, bufsize, stdin);
return *ins ++;
}
template<class T>inline void Read(T &x)
{
int f = 1;
char t = getc();
while (t < '0' || t > '9') {
if (t == '-') f = -1;
t = getc();
}
x = 0;
while (t >= '0' && t <= '9') {
x = x * 10 + t - '0';
t = getc();
}
x *= f;
}
const int maxn = 50005;
struct point {
int P, T; // x, y
point() {}
point(int P, int T) : P(P), T(T) {}
point operator + (const point &rhs) { return point(P + rhs.P, T + rhs.T); }
point operator - (const point &rhs) { return point(P - rhs.P, T - rhs.T); }
point operator * (const double v) { return point(P * v, T * v); }
point operator / (const double v) { return point(P / v, T / v); }
bool operator < (const point &rhs) const { return T * rhs.P < P * rhs.T; }
};
int n, q[maxn], r;
double mx[maxn], mn[maxn], sumu[maxn], sumd[maxn], ratio[maxn];
point g[maxn];
vector<int> ans;
void input()
{
Read(n);
for (register int i = 1; i <= n; i++) Read(g[i].T), Read(g[i].P);
sort(g + 1, g + n + 1);
for (register int i = n; i >= 1; i--) {
sumu[i] = sumu[i + 1] + g[i].T, sumd[i] = sumd[i + 1] + g[i].P;
ratio[i] = sumu[i] / sumd[i];
}
}
inline long long cross(const point &a, const point &b)
{
return a.P * b.T - a.T * b.P;
}
inline double calc(const point &pts, double x)
{
return (double) pts.T - pts.P * x;
}
void solve()
{
// for (register int i = 1; i <= n; i++) printf("%.2f %.2f %.2f\n", g[i].P, g[i].T, g[i].T / g[i].P);
double ratioNow;
r = 0;
for (register int i = 1; i <= n; i++) {
// d = i
ratioNow = ratio[i + 1];
while (r > 0 && g[q[r]].P >= g[i].P) r --;
while (r > 1 && cross(g[i] - g[q[r]], g[q[r]] - g[q[r - 1]]) <= 0) r --;
q[++r] = i;
while (r > 1 && calc(g[q[r]], ratioNow) <= calc(g[q[r - 1]], ratioNow)) r --;
mx[i] = g[q[r]].T - g[q[r]].P * ratioNow;
//printf("d = %d, %.3f %.3f\n", i, ratioNow, mx[i]);
}
r = 0;
for (register int i = n; i >= 2; i--) {
// d = i - 1
ratioNow = ratio[i];
while (r > 0 && g[q[r]].P <= g[i].P) r --;
while (r > 1 && cross(g[q[r]] - g[i], g[q[r - 1]] - g[q[r]]) <= 0) r --;
q[++r] = i;
while (r > 1 && calc(g[q[r]], ratioNow) >= calc(g[q[r - 1]], ratioNow)) r --;
mn[i - 1] = g[q[r]].T - g[q[r]].P * ratioNow;
//printf("d = %d, %.3f %.3f %.3f\n", i - 1, ratioNow, mx[i - 1], mn[i - 1]);
}
for (register int i = 1; i < n; i++) if (mn[i] < mx[i]) ans.push_back(i);
printf("%d\n", (int) ans.size());
for (register int i = 0; i < ans.size(); i++) printf("%d\n", ans[i]);
}
int main()
{
//#ifndef ONLINE_JUDGE
freopen("schul.in", "r", stdin);
freopen("schul.out", "w", stdout);
//#endif
input();
solve();
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
#endif
return 0;
}