我的做法是，枚舉第一個(gè)多邊形的第i條邊和第二個(gè)多邊形的第j條邊重合，然后從這條重合的邊開始，盡可能的向后擴(kuò)展重合邊，然后判斷剩下的多邊形是否是凸多邊形。
比賽的時(shí)候，我在某個(gè)地方忘記對(duì)多邊形點(diǎn)數(shù)求模，導(dǎo)致wa了很久，一直到比賽結(jié)束后才AC。以此為鑒！
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-10-2 12:44:17
File Name: pku3410.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout<<#x<<": "<<x<<endl)
const int maxint=0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template<class T>void show(T a, int n)
{for(int i=0; i<n; ++i) cout<<a[i]<<' '; cout<<endl;}
template<class T>void show(T a, int r, int l){for(int i=0; i<r; ++i)show(a[i],l);cout<<endl;}
const int maxn = 110;
const double eps = 1e-3;
const double pi = acos(-1.0);
typedef struct point_t
{
double x, y;
};
int n1, n2;
point_t p1[maxn], p2[maxn], p[maxn];
int dcmp(double x)
{
if (-eps < x && x < eps)
return 0;
if (x > 0)
return 1;
return -1;
}
double dist2(const point_t &a, const point_t &b)
{
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
double cross(const point_t &a, const point_t &b)
{
return a.x * b.y - a.y * b.x;
}
double cross(const point_t &a1, const point_t &a2, const point_t &b1, const point_t &b2)
{
point_t a, b;
a.x = a2.x - a1.x;
a.y = a2.y - a1.y;
b.x = b2.x - b1.x;
b.y = b2.y - b1.y;
return cross(a, b);
}
double angle(const point_t &a, const point_t &b)
{
double ret = acos((a.x * b.x + a.y * b.y) / (sqrt(a.x * a.x + a.y * a.y) * sqrt(b.x * b.x + b.y * b.y)));
double flag = a.x * b.y - a.y * b.x;
if (flag > 0)
return ret;
else
return -ret;
}
double angle(const point_t &a1, const point_t &a2, const point_t &b1, const point_t &b2)
{
point_t a, b;
a.x = a2.x - a1.x;
a.y = a2.y - a1.y;
b.x = b2.x - b1.x;
b.y = b2.y - b1.y;
return angle(a, b);
}
int main()
{
scanf("%d", &n1);
for (int i = 0; i < n1; i++)
scanf("%lf%lf", &p1[i].x, &p1[i].y);
scanf("%d", &n2);
for (int i = 0; i < n2; i++)
scanf("%lf%lf", &p2[n2 - i - 1].x, &p2[n2 - i - 1].y);
int flag = 0;
for (int i = 0; i < n1; i++)
for (int j = 0; j < n2; j++)
{
int k = 0;
while (1)
{
if (dcmp(dist2(p1[(i + k) % n1], p1[(i + k + 1) % n1]) - dist2(p2[(j + k) % n2], p2[(j + k + 1) % n2])) != 0)
break;
if (k > 0)
if (dcmp(angle(p1[(i + k - 1) % n1], p1[(i + k) % n1], p1[(i + k) % n1], p1[(i + k + 1) % n1])
- angle(p2[(j + k - 1) % n2], p2[(j + k) % n2], p2[(j + k) % n2], p2[(j + k + 1) % n2])) != 0)
break;
k++;
}
if (k == 0)
continue;
if (angle(p1[(n1 + i - 1) % n1], p1[i], p1[i], p1[(i + 1) % n1])
+ angle(p2[(n2 + j - 1) % n2], p2[j], p2[j], p2[(j + 1) % n2]) > pi + eps)
continue;
if (angle(p1[(n1 + i - 1) % n1], p1[i], p1[i], p1[(i + 1) % n1])
+ angle(p2[(n2 + j - 1) % n2], p2[j], p2[j], p2[(j + 1) % n2]) < -eps)
continue;
if (angle(p1[(i + k - 1) % n1], p1[(i + k) % n1], p1[(i + k) % n1], p1[(i + k + 1) % n1])
+ angle(p2[(j + k - 1) % n2], p2[(j + k) % n2], p2[(j + k) % n2], p2[(j + k + 1) % n2]) > pi + eps)
continue;
if (angle(p1[(i + k - 1) % n1], p1[(i + k) % n1], p1[(i + k) % n1], p1[(i + k + 1) % n1])
+ angle(p2[(j + k - 1) % n2], p2[(j + k) % n2], p2[(j + k) % n2], p2[(j + k + 1) % n2]) < -eps)
continue;
int now1 = (i + k) % n1, now2 = (now1 + 1) % n1, now3 = (now2 + 1) % n1;
int flag = 1;
while (now2 != i)
{
if (cross(p1[now1], p1[now2], p1[now2], p1[now3]) < 0)
{
flag = 0;
break;
}
now1 = now2;
now2 = now3;
now3 = (now3 + 1) % n1;
}
if (flag == 0)
continue;
now1 = (j + k) % n2, now2 = (now1 + 1) % n2, now3 = (now2 + 1) % n2;
flag = 1;
while (now2 != j)
{
if (cross(p2[now1], p2[now2], p2[now2], p2[now3]) > 0)
{
flag = 0;
break;
}
now1 = now2;
now2 = now3;
now3 = (now3 + 1) % n2;
}
if (flag == 0)
continue;
printf("1\n");
return 0;
}
printf("0\n");
return 0;
}