• Codeforces Round #524 (Div. 2) C. Masha and two friends 几何:判断矩形是否相交以及相交矩形坐标


    题意 :给出一个初始的黑白相间的棋盘  有两个人  第一个人先用白色染一块矩形区域

    第二个人再用黑色染一块矩形区域 问最后黑白格子各有多少个

    思路:这题的关键在于求相交的矩形区间 

     给出一个矩形的左下和右上角  则相交的条件为 max(X1,X3)<=min(X2,X4)&&max(Y1,Y3)<=min(Y2,Y4)  这里可以用线和线有公共区间的条件来推 矩形就是两个线

    还有就是如果相交则求 相交的矩形的 左下和右上的坐标 ll xa=max(X1,X3),ya=max(Y1,Y3),xb=min(X4,X2),yb=min(Y2,Y4);

    然后就是简单的容斥正常做就行了

     1 #include<bits/stdc++.h>
     2 #define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;i++)
     3 #define MS(arr,arr_value) memset(arr,arr_value,sizeof(arr)) 
     4 #define F first 
     5 #define S second
     6 #define pii pair<int ,int >
     7 #define mkp make_pair
     8 #define pb push_back
     9 #define arr(zzz) array<ll,zzz>
    10 using namespace std;
    11 #define ll long long 
    12 #define int long long 
    13 const int maxn=3e5+5;
    14 const int inf=0x3f3f3f3f;
    15 ll count(int x1,int y1,int x2,int y2,int o){//return black
    16     ll n=x2-x1+1;
    17     ll m=y2-y1+1;
    18     ll white=0,black=0;
    19         if(n&1){
    20             if(m&1){
    21                 white=(n+1)/2+1ll*(n/2+(n+1)/2)*(m/2);
    22                 black=1ll*m*n-white;
    23             }
    24             else {
    25                 white=black=1ll*n*m/2;
    26             }
    27         }
    28         else {
    29             white=black=1ll*n*m/2;
    30         }
    31         if(o==0){
    32             return white;
    33         }
    34         else return black;
    35 
    36 }
    37 void exchange(ll &x,ll&y){
    38     ll tmp=x;
    39     x=y;
    40     y=tmp;
    41 }
    42 int32_t main(){
    43     int t;
    44     scanf("%lld",&t);
    45     int n,m;
    46     while(t--){
    47         ll black=0,white=0;
    48         int X1,Y1,X2,Y2,X3,Y3,X4,Y4;
    49         scanf("%lld%lld",&n,&m);
    50         scanf("%lld%lld%lld%lld%lld%lld%lld%lld",&X1,&Y1,&X2,&Y2,&X3,&Y3,&X4,&Y4);
    51         if(n&1){
    52             if(m&1){
    53                 white=(n+1)/2+  (n/2+(n+1)/2)*(m/2);
    54                 black=1ll*m*n-white;
    55             }
    56             else {
    57                 white=black=1ll*n*m/2;
    58             }
    59         }
    60         else {
    61             white=black=1ll*n*m/2;
    62         }
    63         ll tmp1=count(X1,Y1,X2,Y2,(X1+Y1)%2==0);
    64         white-=1ll*(X2-X1+1)*(Y2-Y1+1)-tmp1;
    65         white+=1ll*(X2-X1+1)*(Y2-Y1+1);
    66         black-=tmp1;
    67         ll tmp2=count(X3,Y3,X4,Y4,(X3+Y3)%2==0);
    68         white-=1ll*(X4-X3+1)*(Y4-Y3+1)-tmp2;
    69         black-=tmp2;
    70         black+=1ll*(X4-X3+1)*(Y4-Y3+1);
    71         /*if(X1<=X3){
    72             exchange(X1,X3);
    73             exchange(X2,X4);
    74             exchange(Y1,Y3);
    75             exchange(Y2,Y4);
    76         }
    77         if(){
    78             
    79         }*/
    80         ll minx=min(X1,X3);
    81         ll miny=min(Y1,Y3);
    82         if(max(X1,X3)<=min(X2,X4)&&max(Y1,Y3)<=min(Y2,Y4)){
    83             ll xa=max(X1,X3),ya=max(Y1,Y3),xb=min(X4,X2),yb=min(Y2,Y4);
    84             ll tmp=count(xa,ya,xb,yb,(xa+ya)%2==0);
    85             //cout<<xa<<" "<<ya<<" "<<xb<<""<<yb<<" fuck"<<tmp<<endl;
    86             black+=tmp;
    87             white-=1ll*(xb-xa+1)*(ya-yb+1);
    88             white+=1ll*(xb-xa+1)*(ya-yb+1)-tmp;
    89             
    90         }
    91          cout<<white<<" "<<black<<endl;
    92     }
    93         
    94     return 0;
    95 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ttttttttrx/p/10793026.html
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