lightoj 1410 - Consistent Verdicts 【思维】

1410 - Consistent Verdicts

PDF (English)

Statistics

Forum

Time Limit: 5 second(s)

Memory Limit: 32 MB

In a 2D plane N persons are standing and each of them has a gun in his hand. The plane is so big that the persons can be considered as points and their locations are given as Cartesian coordinates. Each of the Npersons fire the gun in his hand exactly once and no two of them fire at the same or similar time (the sound of two gun shots are never heard at the same time by anyone so no sound is missed due to concurrency). The hearing ability of all these persons is exactly same. That means if one person can hear a sound at distance R₁, so can every other person and if one person cannot hear a sound at distance R₂ the other N-1 persons cannot hear a sound at distance R₂ as well.

The N persons are numbered from 1 to N. After all the guns are fired, all of them are asked how many gun shots they have heard (not including their own shot) and they give their verdict. It is not possible for you to determine whether their verdicts are true but it is possible for you to judge if their verdicts are consistent. For example, look at the figure above. There are five persons and their coordinates are (1, 2), (3, 1), (5, 1), (6, 3) and (1, 5) and they are numbered as 1, 2, 3, 4 and 5 respectively. After all five of them have shot their guns, you ask them how many shots each of them have heard. Now if there response is 1, 1, 1, 2 and 1 respectively then you can represent it as (1, 1, 1, 2, 1). But this is an inconsistent verdict because if person 4 hears 2 shots then he must have heard the shot fired by person 2, then obviously person 2 must have heard the shot fired by person 1, 3 and 4 (person 1 and 3 are nearer to person 2 than person 4). But their opinions show that Person 2 says that he has heard only 1 shot. On the other hand (1, 2, 2, 1, 0) is a consistent verdict for this scenario so is (2, 2, 2, 1, 1). In this scenario (5, 5, 5, 4, 4) is not a consistent verdict because a person can hear at most 4 shots.

Given the locations of N persons, your job is to find the total number of different consistent verdicts for that scenario. Two verdicts are different if opinion of at least one person is different.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing a positive integer N (1 ≤ N ≤ 700). Each of the next N lines contains two integers x_i y_i (0 ≤ x_i, y_i ≤ 30000) denoting a co-ordinate of a person. Assume that all the co-ordinates are distinct.

Output

For each case, print the case number and the total number of different consistent verdicts for the given scenario.

Sample Input	Output for Sample Input
2 3 1 1 2 2 4 4 2 1 1 5 5	Case 1: 4 Case 2: 2

 

---

PROBLEM SETTER: SHAHRIAR MANZOOR

SPECIAL THANKS: JANE ALAM JAN

题意：给定n个人的坐标，已知每个人的可听距离是相同的。现在n个人依次放枪后，我们询问每个人听到的枪声数目(保证任意两支枪的声音不同时出现)。在所有人回答正确的前提下，有多少种可能存在的答案。

思路：考虑限定每个人的听力距离为D(D为这n个人中某两个人之间的距离)的情况，那么根据这个距离D我们能否唯一确定一组答案？

设出dis(i, j) =  D1 < dis(i, k) = D2，考虑第i个人可以听到的声音数目，在取D2的情况下一定会比取D1时多出一个声音——第k个人传给第i个人。 这样的话，就证明了之前的假设。因此对于这道题，只需要处理出所有距离值，然后去重就可以了。 至于去重，用C++的unique函数 或者 滚动数组都可以实现。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (700+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int dis(int x1, int y1, int x2, int y2){
    return (x1-x2) * (x1-x2) + (y1-y2) * (y1-y2);
}
int x[MAXN], y[MAXN];
int num[MAXN*MAXN];
int main()
{
    int t; Ri(t);
    int kcase = 1;
    W(t)
    {
        int n; Ri(n);
        for(int i = 0; i < n; i++)
            Ri(x[i]), Ri(y[i]);
        int top = 1;
        for(int i = 0; i < n; i++)
            for(int j = i+1; j < n; j++)
                num[top++] = dis(x[i], y[i], x[j], y[j]);
        sort(num+1, num+top);
        int ans = 2;
        for(int i = 2; i < top; i++)
            if(num[i] != num[i-1])
                num[ans++] = num[i];
        printf("Case %d: %d\n", kcase++, ans);
    }
    return 0;
}