我推出来式子为 $\frac{f_{j-1}-f_{k-1}}{b_j-b_k}>-a_i$

于是我维护了个上凸包。

我看了很久都没发现问题，麻烦大佬帮我看看。

#include<bits/stdc++.h>
#define int long long
#define N 70004 
using namespace std;
int n; 
int f[N];
struct node{
	int a, b;
}p[N], c[N]; 
int S[N], tot;
bool cmp(node x, node y) {
	if(x.a == y.a) return x.b < y.b;
	return x.a < y.a;
}
double slope(int k, int j) {
	return 1.000 * (f[j - 1] - f[k - 1]) / (c[j].b - c[k].b);
}
int Q[N], head = 1, tail; //[head, tail]
signed main() {
    scanf("%lld", &n);
    for(int i = 1; i <= n; i++) scanf("%lld %lld", &p[i].a, &p[i].b);
    sort(p + 1, p + n + 1, cmp);
    for(int i = 1; i <= n; i++) {
    	while(tot >= 1 && p[i].b >= p[S[tot]].b) tot--;
    	S[++tot] = i;
	} 
	for(int i = 1; i <= tot; i++) {
		c[i].a = p[S[i]].a;
		c[i].b = p[S[i]].b; 
	}
	n = tot;
	f[1] = c[1].a * c[1].b;
	Q[++tail] = 1;
	for(int i = 2; i <= n; i++) {
		while(head < tail && slope(head, head + 1) > -c[i].a) head++;
		if(head <= tail) {
			int j = Q[head];
			f[i] = min(f[i - 1] + c[i].a * c[i].b, f[j - 1] + c[i].a * c[j].b); 
		}
		while(head < tail && slope(tail, i) > slope(tail - 1, tail)) tail--;
		Q[++tail] = i;
	}
	printf("%lld", f[n]);
	return 0;
}