poj1389:http://poj.org/problem?id=1389
题意：求矩形面积的并
题解：扫描线加线段树 同poj1389
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 const int maxn=2002; 7 int num; 8 struct Node{ 9 int l; 10 int r; 11 int tp; 12 int y; 13 bool operator <(Node a) const{ 14 return y<a.y; 15 } 16 }line[2*maxn]; 17 int arr[2*maxn]; 18 struct Edge{ 19 int left; 20 int right; 21 int flag; 22 int sum; 23 }node1[maxn*4]; 24 void build(int l,int r,int idx){ 25 node1[idx].left=l; 26 node1[idx].right=r; 27 if(l==r){ 28 node1[idx].flag=0; 29 node1[idx].sum=0; 30 return; 31 } 32 int mid=(l+r)/2; 33 build(l,mid,idx<<1); 34 build(mid+1,r,idx<<1|1); 35 node1[idx].sum=node1[idx<<1].sum+node1[idx<<1|1].sum; 36 } 37 void update(int l,int r,int f,int idx){ 38 if(node1[idx].left==node1[idx].right){ 39 node1[idx].flag+=f; 40 if(node1[idx].flag)node1[idx].sum=arr[node1[idx].right+1]-arr[node1[idx].left]; 41 if(!node1[idx].flag)node1[idx].sum=0; 42 return ; 43 } 44 int mid=(node1[idx].left+node1[idx].right)/2; 45 if(mid>=r)update(l,r,f,idx<<1); 46 else if(mid<l)update(l,r,f,idx<<1|1); 47 else{ 48 update(l,mid,f,idx<<1); 49 update(mid+1,r,f,idx<<1|1); 50 } 51 node1[idx].sum=node1[idx<<1].sum+node1[idx<<1|1].sum; 52 } 53 int binaryserach(int x){ 54 int l,r,mid; 55 l=0,r=num+1; 56 while (r-l>1){ 57 mid=(l+r)>>1; 58 if (arr[mid]<=x) l=mid; 59 else r=mid; 60 } 61 return l; 62 } 63 int main(){ 64 int n;int x1,y1,x2,y2; 65 long long ans; 66 while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){ 67 if(x1==-1&&x2==-1&&y1==-1&&y2==-1)break; 68 num=0;int cnt=2; 69 line[1].l=x1;line[1].r=x2;line[1].y=y1;line[1].tp=1; 70 line[2].l=x1;line[2].r=x2;line[2].y=y2;line[2].tp=-1; 71 arr[++num]=x1;arr[++num]=x2; 72 while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){ 73 if(x1==-1&&x2==-1&&y1==-1&&y2==-1)break; 74 75 line[++cnt].l=x1;line[cnt].r=x2;line[cnt].y=y1;line[cnt].tp=1; 76 line[++cnt].l=x1;line[cnt].r=x2;line[cnt].y=y2;line[cnt].tp=-1; 77 arr[++num]=x1;arr[++num]=x2; 78 } 79 80 81 sort(arr+1,arr+num+1); 82 ans=0; 83 sort(line+1,line+cnt+1); 84 build(1,cnt,1); 85 for(int i=1;i<=cnt;i++){ 86 ans+=node1[1].sum*(line[i].y-line[i-1].y); 87 int l=binaryserach(line[i].l); 88 int r=binaryserach(line[i].r)-1; 89 update(l,r,line[i].tp,1); 90 } 91 92 printf("%I64d ",ans); 93 } 94 }