HDU 4562 守护雅典娜 (计算几何+DP)

Posted 2020-10-10 dwtfukgv

题意：略。

析：首先先找出来两种圆，一种是只包含怪物的，另一种是只包含雅典娜的，因为都包含或者都不包含的没什么意义，排序，从小到大按半径，然后分别进行dp，最后肯定是包含的选出最多一个，然后再进行合并。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Point{
  int x, y;
  Point() { }
  Point(int xx, int yy) : x(xx), y(yy) { }
  friend Point operator - (const Point &lhs, const Point &rhs){
    return Point(lhs.x - rhs.x, lhs.y - rhs.y);
  }
};

int dist(Point p){
  return sqr(p.x) + sqr(p.y);
}

struct Circle{
  Point c;
  int r;
  bool operator < (const Circle &c) const{
    return r < c.r;
  }
};
Circle c[maxn];
Point athene, monster;

int isintersection(const Point &p, const Circle &c){
  int d = dist(p - c.c);
  return d - sqr(c.r);
}

bool not_intersection(const Circle &c1, const Circle &c2){
  int d = dist(c1.c - c2.c);
  return d > sqr(c1.r + c2.r);
}

bool isinclude(const Circle &c1, const Circle &c2){
  int d = dist(c1.c - c2.c);
  return sqr(c1.r - c2.r) > d;
}

vector<Circle> aths, mos;
int f[maxn], g[maxn];


int main(){
  monster.x = monster.y = 0;
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    mos.cl; aths.cl;
    scanf("%d %d %d", &n, &athene.x, &athene.y);
    for(int i = 0; i < n; ++i){
      scanf("%d %d %d", &c[i].c.x, &c[i].c.y, &c[i].r);
      int id1 = isintersection(athene, c[i]);
      int id2 = isintersection(monster, c[i]);
      if(id1 < 0 && id2 > 0)  aths.push_back(c[i]);
      else if(id1 > 0 && id2 < 0)  mos.push_back(c[i]);
    }
    sort(aths.begin(), aths.end());
    sort(mos.begin(), mos.end());
    int ans = 0;
    for(int i = 0; i < aths.sz; ++i){
      f[i] = 1;
      for(int j = 0; j < i; ++j)
        if(isinclude(aths[i], aths[j]))  f[i] = max(f[i], f[j] + 1);
      ans = max(ans, f[i]);
    }
    for(int i = 0; i < mos.sz; ++i){
      g[i] = 1;
      for(int j = 0; j < i; ++j)
        if(isinclude(mos[i], mos[j]))  g[i] = max(g[i], g[j] + 1);
      ans = max(ans, g[i]);
    }
    for(int i = 0; i < aths.sz; ++i)
      for(int j = 0; j < mos.sz; ++j)
        if(not_intersection(aths[i], mos[j]))  ans = max(ans, f[i] + g[j]);
    printf("Case %d: %d\n", kase, ans);
  }
  return 0;
}