Warm up 16 Hitting the Targets

Hitting the Targets

Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:65536KB

Total submit users: 38, Accepted users: 38

Problem 12782 : No special judgement

Problem description

又是一道感人的输入方式的傻逼题啊！

A fundamental operation in computational geometry is determining whether two objects touch. For example, in a ame that involves shooting, we want to determine if a player’s shot hits a target. A shot is a two dimensional point, nd a target is a two dimensional enclosed area. A shot hits a target if it is inside the target. The boundary of a target inside the target. Since it is possible for targets to overlap, we want to identify how many targets a shot hits.The figure above illustrates the targets (large unfilled rectangles and circles) and shots (filled circles) of the sample input. The origin (0, 0) is indicated by a small unfilled circle near the center.

Input

Input starts with an integer 1 ≤ m ≤ 30 indicating the number of targets. Each of the next m lines begins with the word rectangle or circle and then a description of the target boundary. A rectangular target’s boundary is given as four integers x1 y1 x2 y2, where x1 < x2 and y1 < y2. The points (x1, y1) and (x2, y2) are the bottom-left and top-right corners of the rectangle, respectively. A circular target’s boundary is given as three integers x y r. The center of the circle is at (x, y) and the 0 < r ≤ 1000 is the radius of the circle. After the target descriptions is an integer 1 ≤ n ≤ 100 indicating the number of shots that follow. The next n lines each contain two integers x y, indicating the coordinates of a shot. All x and y coordinates for targets and shots are in the range [−1000, 1000].

Output

For each of the n shots, print the total number of targets the shot hits.

Sample Input

Sample Input 1
3
rectangle 1 1 10 5
circle 5 0 8
rectangle -5 3 5 8
5
1 1
4 5
10 10
-10 -1
4 -3

Sample Output

Sample Output 1
2
3
0
0
1

Problem Source

HNU Contest 

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#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1305
#define ll long long
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
int n, m;
ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
char s[maxn][maxn];
struct rec{int x1, x2, y1, y2;}re[maxn];
struct cir{int x, y, r;}ci[maxn];
int main(){
    /*int t;
    scanf("%d", &t);
    while (t--){
    scanf("%I64d", &n);

    }*/
    //ll x, y, x1, x2, y1, y2, s1, s2;
    while (~scanf("%d", &n)){
        for (int i = 0; i < n; i++){
            scanf("%s", s[i]);
            if (s[i][0] == 'r')scanf("%d%d%d%d", &re[i].x1, &re[i].y1, &re[i].x2, &re[i].y2);
            if (s[i][0] == 'c')scanf("%d%d%d", &ci[i].x, &ci[i].y, &ci[i].r);
        }
        scanf("%d", &m);
        while (m--){
            int x, y;
            scanf("%d%d", &x, &y);
            int t = 0;
            for (int i = 0; i < n; i++){
                if (s[i][0] == 'r')if (x >= re[i].x1&&x <= re[i].x2&&y >= re[i].y1&&y <= re[i].y2)t++;
                else {
                    double a, b, dx, dy, dr;
                    a = x; b = y; dx = ci[i].x, dy = ci[i].y; dr = ci[i].r;
                    if (sqrt((dx - x)*(dx - x) + (dy - y)*(dy - y)) <= dr)t++;
                }
            }
            printf("%d\n", t);
        }
    }
    return 0;
}