題目要求從幾個(gè)區(qū)域中，求出包含其它區(qū)域的那個(gè)區(qū)域。其實(shí)就是求最大區(qū)域。
只要對(duì)每個(gè)區(qū)域依次計(jì)算面積即可，然后取最大的那個(gè)。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-13 14:45:09
File Name: pku1687.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
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 = 100;
const double eps = 1e-9;
const double pi = acos(-1.0);
typedef struct point_t
{
int x, y;
};
typedef struct polygon_t
{
int n;
point_t p[maxn];
};
int64 det(const point_t &a, const point_t &b)
{
return int64(a.x) * b.y - int64(a.y) * b.x;
}
point_t p[maxn];
int n;
int main()
{
int ca;
for (scanf("%d", &ca); ca--;)
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d", &p[i].x, &p[i].y);
int m;
scanf("%d", &m);
int64 max_area = 0;
int ans;
for (int i = 0; i < m; i++)
{
polygon_t poly;
scanf("%d", &poly.n);
for (int j = 0; j < poly.n; j++)
{
int t;
scanf("%d", &t);
poly.p[j] = p[t - 1];
}
int64 area = 0;
for (int j = 0; j < poly.n; j++)
area += det(poly.p[j], poly.p[(j + 1) % poly.n]);
if (abs(area) > max_area)
{
max_area = abs(area);
ans = i + 1;
}
}
printf("%d\n", ans);
}
return 0;
}