题意：二维平面上有两个框，问平面被分成了几个部分
x，y<=1e9
思路：分类讨论可以
但数据范围实在太小了，离散化以后随便dfs一下
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 #define N 110000 11 #define M 4100000 12 #define fi first 13 #define se second 14 #define MP make_pair 15 #define pi acos(-1) 16 #define mem(a,b) memset(a,b,sizeof(a)) 17 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 18 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 19 #define lowbit(x) x&(-x) 20 #define Rand (rand()*(1<<16)+rand()) 21 #define id(x) ((x)<=B?(x):m-n/(x)+1) 22 #define ls p<<1 23 #define rs p<<1|1 24 25 const ll MOD=1e9+7,inv2=(MOD+1)/2; 26 double eps=1e-6; 27 int INF=1e9; 28 int dx[4]={-1,1,0,0}; 29 int dy[4]={0,0,-1,1}; 30 31 int read() 32 { 33 int v=0,f=1; 34 char c=getchar(); 35 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 36 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 37 return v*f; 38 } 39 40 struct arr 41 { 42 int x,id; 43 }x[10],y[10]; 44 int b[20][20],c[20],d[20]; 45 46 bool cmp(arr a,arr b) 47 { 48 return a.x<b.x; 49 } 50 51 void dfs(int x,int y) 52 { 53 //printf("x=%d y=%d ",x,y); 54 b[x][y]=1; 55 rep(i,0,3) 56 { 57 int ux=x+dx[i],uy=y+dy[i]; 58 if(ux>0&&ux<15&&uy>0&&uy<15&&b[ux][uy]==0) dfs(ux,uy); 59 } 60 } 61 62 int main() 63 { 64 //freopen("1.in","r",stdin); 65 int cas; 66 scanf("%d",&cas); 67 while(cas--) 68 { 69 x[1].x=read(),y[1].x=read(),x[2].x=read(),y[2].x=read(); 70 x[3].x=read(),y[3].x=read(),x[4].x=read(),y[4].x=read(); 71 rep(i,1,4) 72 { 73 x[i].id=y[i].id=i; 74 } 75 sort(x+1,x+4+1,cmp); 76 sort(y+1,y+4+1,cmp); 77 int m1=2,m2=2; 78 c[x[1].id]=2; 79 d[y[1].id]=2; 80 rep(i,2,4) 81 { 82 if(x[i].x==x[i-1].x) c[x[i].id]=m1; 83 else 84 { 85 m1+=2; 86 c[x[i].id]=++m1; 87 } 88 if(y[i].x==y[i-1].x) d[y[i].id]=m2; 89 else 90 { 91 m2+=2; 92 d[y[i].id]=++m2; 93 } 94 } 95 rep(i,1,15) 96 rep(j,1,15) b[i][j]=0; 97 rep(i,d[1],d[2]) b[c[1]][i]=b[c[2]][i]=1; 98 rep(i,d[3],d[4]) b[c[3]][i]=b[c[4]][i]=1; 99 rep(i,c[1],c[2]) b[i][d[2]]=b[i][d[1]]=1; 100 rep(i,c[3],c[4]) b[i][d[4]]=b[i][d[3]]=1; 101 int ans=0; 102 rep(i,1,14) 103 rep(j,1,14) 104 if(b[i][j]==0) 105 { 106 ans++; 107 dfs(i,j); 108 } 109 110 printf("%d ",ans); 111 112 113 } 114 115 return 0; 116 }