这道题比赛之后被重新加了几个case,很多人现在都过不了了
算法就是先求凸包,然后判断两个凸包相等
我们可以吧凸包序列化为两点距离和角度
角度如果直接拿向量的叉积是不对的,,因为钝角和锐角的叉积有可能相同。我直接把点积和叉积加一起当作角度其实也不严谨,,最好是变成三个元素,长度,叉积,点积
代码有所参考
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <functional>
#include <iomanip>
using namespace std;
#define LL long long
const int maxn = 100000 + 100;
struct Point {
LL x, y;
Point() {}
Point(LL xx, LL yy) {
x = xx;
y = yy;
}
};
LL operator*(const Point &a, const Point &b) {
return a.x * b.x + a.y * b.y;
}
bool operator<(const Point &a, const Point &b) {
return a.x == b.x? a.y < b.y: a.x < b.x;
}
LL operator^(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
Point operator-(const Point &a, const Point &b) {
return Point(a.x - b.x, a.y - b.y);
}
struct Cross_len {
LL cross, len;
};
bool operator==(const Cross_len &a, const Cross_len &b) {
return a.cross == b.cross && a.len == b.len;
}
bool operator!=(const Cross_len &a, const Cross_len &b) {
return !(a == b);
}
int n[2], top[2];
Point point[2][maxn], sta[2][maxn << 1];
int Next[maxn];
Cross_len cross_len[2][maxn << 1];
void ConvexHull(Point *p, int n, int &top, Point *sta) {
top = 0;
sort(p + 1, p + 1 + n);
for(int i = 1; i <= n; ++i) {
while(top > 1 && ((sta[top - 1] - sta[top - 2]) ^ (p[i] - sta[top - 2])) <= 0) {
--top;
}
sta[top++] = p[i];
}
int ttop = top;
for(int i = n; i > 0; --i) {
while(top > ttop && ((sta[top - 1] - sta[top - 2]) ^ (p[i] - sta[top - 2])) <= 0) {
--top;
}
sta[top++] = p[i];
}
--top;
}
void Change_to_Cross_len(Point *p, int n, Cross_len *c) {
// for(int i = 0; i < n; ++i) {
// printf("%lld %lld / ", p[i].x, p[i].y);
// }
// printf("
");
Point p0 = p[0];
for(int i = 0; i < n - 1; ++i) {
p[i] = p[i + 1] - p[i];
c[i].len = p[i] * p[i];
}
p[n - 1] = p0 - p[n - 1];
c[n - 1].len = p[n - 1] * p[n - 1];
for(int i = 0; i < n; ++i) {
c[i].cross = (p[i] * p[(i + 1) % n]) + (p[i] ^ p[(i + 1) % n]);
}
// for(int i = 0; i < n; ++i) {
// printf("%lld %lld /%lld %lld: ", p[i].x, p[i].y, c[i].cross, c[i].len);
// }
// printf("
");
}
void get_next(Cross_len *str, int n) {
int j = -1;
Next[0] = -1;
for(int i = 1; i < n; ++i) {
while(j != -1 && str[j + 1] != str[i]) {
j = Next[j];
}
if(str[i] == str[j + 1]) {
++j;
}
Next[i] = j;
}
}
bool kmp(Cross_len *str1, int n, Cross_len *str2, int m) {
int j = -1;
for(int i = 0; i < n; ++i) {
while(j != -1 && str1[i] != str2[j + 1]) {
j = Next[j];
}
if(str1[i] == str2[j + 1]) {
++j;
}
if(j == m - 1) {
return true;
}
}
return false;
}
int main() {
while(scanf("%d%d", &n[0], &n[1]) != EOF) {
memset(point, 0, sizeof(point));
for(int i = 0; i < 2; ++i) {
for(int j = 1; j <= n[i]; ++j) {
scanf("%lld%lld", &point[i][j].x, &point[i][j].y);
}
ConvexHull(point[i], n[i], top[i], sta[i]);
if(i == 0) {
for(int j = 0; j < top[i]; ++j) {
sta[i][j + top[i]] = sta[i][j];
}
top[i] *= 2;
}
Change_to_Cross_len(sta[i], top[i], cross_len[i]);
}
get_next(cross_len[1], top[1]);
if(kmp(cross_len[0], top[0], cross_len[1], top[1])) {
printf("Yes
");
} else {
printf("No
");
}
}
return 0;
}