UVa 1325 - Hypertransmission (思维 + 扫描)
题意
每个星球可以收到两种广播,一种是自己人,一种是敌对势力。现在敌对势力想使星球收到的敌对广播大于自己人广播,并且使这样的星球数量最大。
在数量一样大的时候,广播的距离越小越好。
现在求最大数量和前提下的最小距离。
思路
答案一定在星球的两两距离之中。
所以可以求出两两距离,排序,从小的开始统计。因为半径加大,之前的答案一定是满足的。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
LL INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1000 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
struct POINT
{
int x, y, z, type, cnt;
}p[MAXN];
struct EDGE
{
int l, r, len;
bool operator < (const EDGE &a) const
{
return len < a.len;
}
};
int Dis(POINT a, POINT b)
{
return (a.x-b.x) * (a.x-b.x) + (a.y-b.y) * (a.y-b.y) + (a.z-b.z) * (a.z-b.z);
}
vector<EDGE>edges;
int main()
{
//ROP;
int n;
while (~scanf("%d", &n))
{
edges.clear();
int a, b, c, type;
for (int i = 0; i < n; i++)
{
scanf("%d%d%d%d", &p[i].x, &p[i].y, &p[i].z, &p[i].type);
p[i].cnt = 1;
}
for (int i = 0; i < n; i++)
for (int j = i+1; j < n; j++) edges.PB((EDGE){i, j, Dis(p[i], p[j])});
sort(edges.begin(), edges.end());
int tmp = 0, i, j;
pair<int, double> ans;
for (i = 0; i < SZ(edges); )
{
for (j = i; j < SZ(edges) && edges[i].len == edges[j].len; j++)
{
int a = edges[j].l, b = edges[j].r;
if (p[a].type != p[b].type)
{
if (--p[a].cnt == -1) tmp++;
if (--p[b].cnt == -1) tmp++;
}
else
{
if (++p[a].cnt == 0) tmp--;
if (++p[b].cnt == 0) tmp--;
}
}
if (tmp > ans.X)
ans.X = tmp, ans.Y = edges[i].len;
i = j;
}
printf("%d\n%.6f\n", ans.X, sqrt(ans.Y));
}
return 0;
}