扫描线求面积的并：题目
模板：
1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4 typedef long long ll; 5 const int maxn = 210; 6 int n; 7 double x[maxn<<2]; 8 struct node{ 9 double l,r,h; 10 int flag; 11 }nd[maxn<<2]; 12 13 bool cmp(node u, node v) 14 { 15 return u.h<v.h; 16 } 17 18 struct tree{ 19 int l,r,cnt; 20 double len; 21 }tree[maxn<<2]; 22 23 void build(int rt,int left,int right){ 24 tree[rt].l=left; 25 tree[rt].r=right; 26 tree[rt].len=0; 27 tree[rt].cnt=0; 28 if( left==right ) return ; 29 int mid=(left+right)>>1; 30 build(rt<<1,left,mid); 31 build(rt<<1|1,mid+1,right); 32 } 33 34 int findPos(int l,int r,double val){ 35 int mid; 36 while( l<=r ){ 37 mid=(l+r)>>1; 38 if( x[mid]>val ) r=mid-1; 39 else if( x[mid]<val ) l=mid+1; 40 else break; 41 } 42 return mid; 43 } 44 45 void pushUp(int rt){ 46 if( tree[rt].cnt ){//非0,整段覆盖 47 tree[rt].len=x[tree[rt].r+1]-x[tree[rt].l]; 48 } 49 else if( tree[rt].l==tree[rt].r )//叶子 50 tree[rt].len=0; 51 else tree[rt].len=tree[rt<<1].len+tree[rt<<1|1].len; 52 } 53 54 void update(int rt,int left,int right,int val){ 55 if( left<=tree[rt].l && tree[rt].r<=right ){ 56 tree[rt].cnt+=val; 57 pushUp(rt); 58 return ; 59 } 60 int mid=(tree[rt].l+tree[rt].r)>>1; 61 if( left<=mid ) update(rt<<1,left,right,val); 62 if( right>mid ) update(rt<<1|1,left,right,val); 63 pushUp(rt); 64 } 65 66 int main() 67 { 68 int cas=0; 69 int l,r; 70 double x1,x2,y1,y2; 71 while( ~scanf("%d",&n)&&n){ 72 int cnt=0; 73 for(int i=1;i<=n;i++){ 74 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 75 x[++cnt]=x1; 76 nd[cnt].l=x1; 77 nd[cnt].r=x2; 78 nd[cnt].h=y1; 79 nd[cnt].flag=1;//下边 80 x[++cnt]=x2; 81 nd[cnt].l=x1; 82 nd[cnt].r=x2; 83 nd[cnt].h=y2; 84 nd[cnt].flag=-1;//上边 85 } 86 sort(x+1,x+1+cnt); 87 sort(nd+1,nd+1+cnt,cmp); 88 build(1,1,cnt); 89 double ans=0; 90 for(int i=1;i<=cnt;i++){ 91 l=findPos(1,cnt,nd[i].l); 92 r=findPos(1,cnt,nd[i].r)-1; 93 update(1,l,r,nd[i].flag); 94 ans += tree[1].len*(nd[i+1].h-nd[i].h); 95 } 96 printf("Test case #%d Total explored area: %.2f ",++cas,ans); 97 } 98 return 0; 99 }
扫描线先求面积的交：题目
模板：
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 typedef long long ll; 7 8 const int maxn=2010; 9 int n, cnt; 10 double x[maxn<<2]; 11 struct Node{ 12 double l,r,y; 13 int flag; 14 }node[maxn<<2]; 15 16 int cmp(Node a ,Node b){ 17 return a.y<b.y; 18 } 19 20 struct Tree{ 21 int l, r; 22 double add,len1,len2; 23 }tree[maxn<<2]; 24 25 int findPos(double val)//找val在x[]中出现的位置 26 { 27 int l=1, r=cnt; 28 int mid; 29 while(l<=r) 30 { 31 mid=(l+r)>>1; 32 if(x[mid]<val) l=mid+1; 33 else if(x[mid]>val) r=mid-1; 34 else break; 35 } 36 return mid; 37 } 38 39 void build(int rt, int left, int right) 40 { 41 tree[rt].l=left; 42 tree[rt].r=right; 43 tree[rt].add=tree[rt].len1=tree[rt].len2=0;//全置0 44 if(left==right)//叶子 45 { 46 return ; 47 } 48 int mid=(left+right)>>1; 49 build(rt<<1,left,mid); 50 build(rt<<1|1,mid+1,right); 51 } 52 void pushUp(int rt) 53 { 54 if(tree[rt].add>=2) tree[rt].len1=tree[rt].len2=x[tree[rt].r+1]-x[tree[rt].l]; 55 else if(tree[rt].add==1){ 56 tree[rt].len1=x[tree[rt].r+1]-x[tree[rt].l]; 57 if(tree[rt].l==tree[rt].r) tree[rt].len2=0; 58 else tree[rt].len2=tree[rt<<1].len1+tree[rt<<1|1].len1; 59 } 60 else{ 61 if(tree[rt].l==tree[rt].r) tree[rt].len1=tree[rt].len2=0; 62 else{ 63 tree[rt].len1=tree[rt<<1].len1+tree[rt<<1|1].len1; 64 tree[rt].len2=tree[rt<<1].len2+tree[rt<<1|1].len2; 65 } 66 } 67 } 68 69 void update(int rt,int left,int right,int val){ 70 if(left<=tree[rt].l && tree[rt].r<=right){ 71 tree[rt].add+=val; 72 pushUp(rt); 73 return ; 74 } 75 int mid=(tree[rt].l+tree[rt].r)>>1; 76 if( left<=mid ) update(rt<<1,left,right,val); 77 if( right>mid ) update(rt<<1|1,left,right,val); 78 pushUp(rt); 79 } 80 81 int main() 82 { 83 double x1,x2,y1,y2; 84 int l,r,t; 85 scanf("%d",&t); 86 while(t--){ 87 cnt=0; 88 memset(node, 0, sizeof(node)); 89 memset(tree, 0, sizeof(tree)); 90 scanf("%d",&n); 91 for(int i=1;i<=n;i++){ 92 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 93 x[++cnt]=x1; 94 node[cnt].l=x1; 95 node[cnt].r=x2; 96 node[cnt].y=y1; 97 node[cnt].flag=1;//左边 98 x[++cnt]=x2; 99 node[cnt].l=x1; 100 node[cnt].r=x2; 101 node[cnt].y=y2; 102 node[cnt].flag=-1;//右边 103 } 104 sort(x+1,x+cnt+1); 105 sort(node+1,node+cnt+1,cmp); 106 build(1,1,cnt); 107 double ans=0; 108 109 l=findPos(node[1].l); 110 r=findPos(node[1].r)-1; 111 update(1,l,r,node[1].flag); 112 for(int i=2;i<=2*n;i++){ 113 ans+=tree[1].len2*(node[i].y-node[i-1].y); 114 l=findPos(node[i].l); 115 r=findPos(node[i].r)-1; 116 update(1, l, r, node[i].flag); 117 } 118 printf("%.2f ", ans); 119 } 120 return 0; 121 }