http://acm.hdu.edu.cn/showproblem.php?pid=1255
这题...如果会求面积并的话，面积交就不是问题了。只需要将原先判断线段树中cover值为大于等于1改为大于等于2即可！
什么都不会的同学，请看上一个随笔http://www.cnblogs.com/ka200812/archive/2011/11/13/2247064.html
View Code
1 #include<iostream>
2 #include<string>
3 #include<stdlib.h>
4 using namespace std;
5
6 typedef struct
7 {
8 double x;
9 double y_up;
10 double y_down;
11 int l_r;
12 }LINE;
13
14 typedef struct
15 {
16 double x;
17 double y_up;
18 double y_down;
19 int cover;
20 bool has_child;
21 }TREE;
22
23 int cmp(const void *a,const void *b)
24 {
25 return *(double *)a>*(double *)b? 1: -1;
26 }
27
28 TREE tree[1000*1001];
29 LINE line[2002];
30 int n,index;
31 double x1,y1,x2,y2;
32 double y[2002];
33
34 void build(int i,int l,int r)
35 {
36 tree[i].cover=0;
37 tree[i].x=-1;
38 tree[i].has_child=true;
39 tree[i].y_up=y[r];
40 tree[i].y_down=y[l];
41 if(l+1==r)
42 {
43 tree[i].has_child=false;
44 return;
45 }
46 int mid=(l+r)>>1;
47 build(2*i,l,mid);
48 build(2*i+1,mid,r);
49 }
50
51 double insert(int i,double l,double r,double x,int l_r)
52 {
53 if(tree[i].y_up<=l || tree[i].y_down>=r)
54 return 0;
55 if(!tree[i].has_child)
56 {
57 if(tree[i].cover>1)
58 {
59 double ans=(x-tree[i].x)*(tree[i].y_up-tree[i].y_down);
60 tree[i].x=x;
61 tree[i].cover+=l_r;
62 return ans;
63 }
64 else
65 {
66 tree[i].cover+=l_r;
67 tree[i].x=x;
68 return 0;
69 }
70 }
71 else
72 return insert(2*i,l,r,x,l_r)+insert(2*i+1,l,r,x,l_r);
73 }
74
75 int main()
76 {
77 int t,i;
78 //freopen("D:\\in.txt","r",stdin);
79 cin>>t;
80 while(t--)
81 {
82 cin>>n;
83 index=1;
84 for(i=1;i<=n;i++)
85 {
86 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
87 line[index].x=x1;
88 line[index].l_r=1;
89 line[index].y_up=y2;
90 line[index].y_down=y1;
91 y[index]=y1;
92 index++;
93
94 line[index].x=x2;
95 line[index].l_r=-1;
96 line[index].y_up=y2;
97 line[index].y_down=y1;
98 y[index]=y2;
99 index++;
100 }
101 qsort(y+1,2*n,sizeof(y[0]),cmp);
102 qsort(line+1,2*n,sizeof(line[0]),cmp);
103 build(1,1,index-1);
104 double ans=0;
105 for(i=1;i<index;i++)
106 {
107 ans+=insert(1,line[i].y_down,line[i].y_up,line[i].x,line[i].l_r);
108 }
109 printf("%.2lf\n",ans);
110 }
111 return 0;
112 }