ZR#317.【18 提高 2】A(计算几何 二分)

题意

 

sol

非常好的一道题，幸亏这场比赛我没打，不然我估计要死在这个题上qwq

到不是说有多难，关键是细节太多了，我和wcz口胡了一下我的思路，然后他写了一晚上没调出来qwq

解法挺套路的，先提出一个$x$

然后维护一堆直线对应的上凸壳

在凸壳上二分即可。

由于这题的$x$很小，直接处理出答案就行了

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define ull unsigned long long 
#define rg register 
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? eof : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *o = obuf;
//void print(int x) {if(x > 9) print(x / 10); *o++ = x % 10 + '0';}
//#define os  *o++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int maxn = 1e6 + 10, inf = 1e9 + 10, mod = 1e9 + 7, bb = 32323;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n;
struct node {
    double a, b;
    bool operator < (const node &rhs) const {
        return a == rhs.a ? b < rhs.b : a < rhs.a;
    }
}p[maxn], s1[maxn], s2[maxn];
int t1 = 0, t2 = 0, ans[maxn];
double cross(node x, node y) {
//    printf("%lf\n", 1.0 * (y.b - x.b) / (x.a - x.b));
    return 1.0 * (y.b - x.b) / (x.a - y.a);
}
void get() {
    sort(p + 1, p + n + 1);
    s1[++t1] = p[1];
    for(int i = 2; i <= n; i++) {
        if(t1 && p[i].a == s1[t1].a) t1--;
        while(t1 > 1 && cross(p[i], s1[t1]) <= cross(s1[t1], s1[t1 - 1])) t1--;
        s1[++t1] = p[i];
    }
    for(int i = 1; i <= n; i++) p[i].b = -p[i].b;
    sort(p + 1, p + n + 1);
    s2[++t2] = p[1];
    for(int i = 2; i <= n; i++) {
        if(t1 && p[i].a == s2[t2].a) t2--;
        while(t2 > 1 && cross(p[i], s2[t2]) <= cross(s2[t2], s2[t2 - 1])) t2--;
        s2[++t2] = p[i];
    }
}
int query(node p, int x) {
    return p.a * x * x + p.b * x;
}
void makeans() {
    for(int i = 1, c = 1; i <= bb; i++) {
        //printf("%lf\n", cross(s1[c], s1[c + 1]));
        while(c < t1 && cross(s1[c], s1[c + 1]) <= (double)i) c++;
        ans[i + bb] = query(s1[c], i);
    }
    for(int i = 1, c = 1; i <= bb; i++) {
        while(c < t2 && cross(s2[c], s2[c + 1]) <= (double)i) c++;
        ans[bb - i] = query(s2[c], i);
    }
}
main() {
//    freopen("a.in", "r", stdin);
    n = read(); int q = read();
    for(int i = 1; i <= n; i++) p[i].a = read(), p[i].b = read();
    get();
    makeans();
    while(q--) {
        int x = read();
        printf("%lld\n", ans[x + bb]);
    }
    return 0;
}
/*
2 2 1
1 1
2 1 1
*/