ZR#317.【18 提高 2】A(计算几何 二分)
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2022-04-14 21:57:32
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 ......
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 */
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