CERC17 Problem L - Lunar Landscape

Solution

如果只有$A$，我们可以简单地使用差分解决。
加入了$B$之后，可以把单位正方形用对角线拆成四个等腰Rt，统计等腰Rt的个数。我们先进行坐标转换，把$(x,y)->(x-y,x+y)$，然后差分。

时间复杂度$O(n+W^2)$。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
const int L=1501;
int A[L*2][L*2],B[L*4][L*4];
signed main()
{int n=read();for (int i=1;i<=n;i++){char c; scanf("%c",&c);int x=read(),y=read(),z=read()>>1;if (c=='A') x+=L,y+=L,A[x-z][y-z]++,A[x-z][y+z]--,A[x+z][y-z]--,A[x+z][y+z]++;else {int X=x-y+L*2,Y=x+y+L*2;B[X-z][Y-z]++,B[X-z][Y+z]--,B[X+z][Y-z]--,B[X+z][Y+z]++;}}for (int i=1;i<L*2;i++) for (int j=0;j<L*2;j++) A[i][j]+=A[i-1][j];for (int i=0;i<L*2;i++) for (int j=1;j<L*2;j++) A[i][j]+=A[i][j-1];for (int i=1;i<L*4;i++) for (int j=0;j<L*4;j++) B[i][j]+=B[i-1][j];for (int i=0;i<L*4;i++) for (int j=1;j<L*4;j++) B[i][j]+=B[i][j-1];int ans=0;for (int i=-L;i<L;i++)for (int j=-L;j<L;j++)if (A[i+L][j+L]) ans+=4;else ans+=(B[i-j+L*2][i+j+L*2]>0)+(B[i-j+L*2-1][i+j+L*2]>0)+(B[i-j+L*2][i+j+L*2+1]>0)+(B[i-j+L*2-1][i+j+L*2+1]>0);printf("%.2lf\n",ans*.25);return 0;
}