+ Scan line segment tree is solved by a rectangular footprint required area / perimeter problem
Area edition:
That is, given a number of rectangles, and finally find the total area (the focus is quickly resolve coverage issues)
A rectangular footprint
Three rectangles stacked together will produce duplicate part, to how to solve this problem?
Such problems are generally used tree line in the auxiliary scanning method to calculate!
Scanning line as shown below, is obtained as long as the effective length of each scan line can be drawn on the actual area of the region, and finally summing all the area, is the total area of the polygon.
Scan line
Gangster address: https: //www.jianshu.com/p/4b71c60e290b
capacity 1151
AC:
1 #include<iostream> 2 #include<cstring> 3 //#include<bits/stdc++.h> 4 #include<math.h> 5 #include<algorithm> 6 #include<queue> 7 #include<stack> 8 #include<cstdio> 9 #include<map> 10 #include<set> 11 #define si(a) scanf("%d",&a) 12 #define sl(a) scanf("%lld",&a) 13 #define sii(a,b) scanf("%d%d",&a,&b) 14 #define sll(a,b) scanf("%lld%lld",&a,&b) 15 #define queues priority_queue 16 #define mod 1000000007 17 #define mem(a) memset(a,0,sizeof(a)); 18 #define def(a) ((a)&(-a)) 19 #define fi first 20 #define se second 21 #define mp make_pair 22 #define pb push_back 23 #define mem(a,b) memset(a,b,sizeof(a)); 24 typedef long long ll; 25 //priority_queue<ll,vector<ll >,greater<ll > >q; 26 const int INF=0x3f3f3f3f; 27 //const double E=exp(1); 28 //const double PI=acos(-1); 29 using namespace std; 30 //__builtin_popcount 31 //priority_queue 32 const ll MAX=100000; 33 //ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);} 34 //ll kc(ll a,ll b){ll c=1;while(b){if(b&1)c=(c*a)%mod;a=(a*a)%mod;b>>=1;}return c;} 35 //bool cmp(double x,double y){return x>y;} 36 //ll F[50003*3]; 37 //ll Find(ll a){return F[a]<0?a:F[a]=Find(F[a]);} 38 //ll same(ll a,ll b){if(Find(a)==Find(b))return 1;return 0;} 39 //ll up1(ll a,ll b){ll x,y;x=Find(a);y=Find(b);if(x!=y)F[x]=y;return 0;} 40 //int read(){int x=0,f=1;char s=getchar();for(; s>'9'||s<'0'; s=getchar()) if(s=='-') f=-1;for(; s>='0'&&s<='9'; s=getchar()) x=x*10+s-'0';return x*f;} 41 struct Node 42 { 43 double xl,xr,y; 44 int s; 45 Node(double x,double x1,double y1,int s1):xl(x),xr(x1),y(y1),s(s1) {} 46 bool operator <(Node z) 47 { 48 return y<z.y; 49 } 50 Node() {} 51 } node[203]; 52 struct Tree 53 { 54 int tl,tr; 55 int s; 56 double len; 57 } tree[203<<2]; 58 double X[203],X1[203]; 59 int m; 60 void build(int id,int l,int r)//初始化 61 { 62 tree[id].tl=l; 63 tree[id].tr=r; 64 tree[id].s=0; 65 tree[id].len=0; 66 if(l==r)return ; 67 int mid=(l+r)>>1; 68 build(id<<1,l,mid); 69 build(id<<1|1,mid+1,r); 70 } 71 int b=0; 72 int c; 73 void pushup(int id)//更新 74 { 75 if(tree[id].s) 76 { 77 tree[id].len=X[tree[id].tr+1]-X[tree[id].tl]; 78 } 79 else if(tree[id].tr==tree[id].tl) tree[id].len=0; 80 else 81 { 82 tree[id].len=tree[id<<1].len+tree[id<<1|1].len; 83 } 84 } 85 86 void update(int id,int l,int r,int v) 87 { 88 int tl; 89 int tr; 90 tl=tree[id].tl; 91 tr=tree[id].tr; 92 if(l<=tl&&tr<=r) 93 { 94 tree[id].s+=v; 95 pushup(id); 96 return ; 97 } 98 int mid=(tr+tl)>>1; 99 if(mid>=l) update(id<<1,l,r,v); 100 if(mid<r)update(id<<1|1,l,r,v); 101 pushup(id); 102 103 } 104 int main() 105 { 106 // ios::sync_with_stdio(false); 107 int t; 108 int ant=0; 109 while(scanf("%d",&t)&&t) 110 { 111 m=0; 112 for(int i=1; i<=t; i++) 113 { 114 double x1,x2,y1,y2; 115 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 116 node[++m]=Node(x1,x2,y1,1); 117 X1[m]=x1; 118 node[++m]=Node(x1,x2,y2,-1); 119 X1[m]=x2; 120 } 121 sort(X1+1,X1+m+1); 122 sort(node+1,node+1+m); 123 double are=0; 124 int mm=0; 125 X[++mm]=X1[1]; 126 for(int i=2; i<=m; i++) //去重 127 if(X1[i]!=X1[i-1]) 128 X[++mm]=X1[i]; 129 build(1,1,mm); 130 for(int i=1; i<=m; i++) 131 { 132 int l=lower_bound(X+1,X+1+mm,node[i].xl)-X; 133 int r=lower_bound(X+1,X+1+mm,node[i].xr)-X-1; 134 c=i; 135 update(1,l,r,node[i].s); 136 are+=tree[1].len*(node[i+1].y-node[i].y); 137 } 138 printf("Test case #%d\n",++ant); 139 printf("Total explored area: %.2f\n\n",are); 140 } 141 }