因为要求覆盖两次的面积，所以可以开二维表示被覆盖1次及以上和被覆盖两次及以上的面积。
主要在up更新上，如果当前区间被标记为一次，二次覆盖的面积就为其儿子面积之和，具体看代码吧。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<map> 11 using namespace std; 12 #define N 4010 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 map<double,int>f; 19 double s[N<<2],res[N<<2]; 20 int fs[N<<2]; 21 double val[N],que[N]; 22 struct node 23 { 24 double x1,x2,y; 25 int f; 26 node(){} 27 node(double x1,double x2,double y,int f):x1(x1),x2(x2),y(y),f(f){} 28 bool operator < (const node &s) const 29 { 30 return y <s.y; 31 } 32 }p[N]; 33 void up(int w,int l,int r) 34 { 35 if(fs[w]>1) 36 { 37 s[w] = val[r+1]-val[l]; 38 res[w] = s[w]; 39 } 40 else if(fs[w]==1) 41 { 42 if(l==r) 43 s[w] = 0; 44 else 45 s[w] = res[w<<1]+res[w<<1|1]; 46 res[w] = val[r+1]-val[l]; 47 } 48 else 49 { 50 if(l==r) 51 { 52 s[w] = 0; 53 res[w] = 0; 54 } 55 else 56 { 57 s[w] = s[w<<1]+s[w<<1|1]; 58 res[w] = res[w<<1]+res[w<<1|1]; 59 } 60 } 61 } 62 void build(int l,int r,int w) 63 { 64 fs[w] = 0; 65 res[w] = s[w] = 0; 66 if(l==r) 67 { 68 return ; 69 } 70 int m = (l+r)>>1; 71 build(l,m,w<<1); 72 build(m+1,r,w<<1|1); 73 up(w,l,r); 74 } 75 void update(int a,int b,int d,int l,int r,int w) 76 { 77 if(a<=l&&b>=r) 78 { 79 fs[w]+=d; 80 up(w,l,r); 81 return ; 82 } 83 int m = (l+r)>>1; 84 if(a<=m) update(a,b,d,l,m,w<<1); 85 if(b>m) update(a,b,d,m+1,r,w<<1|1); 86 up(w,l,r); 87 } 88 int main() 89 { 90 int t,i,n; 91 scanf("%d",&t); 92 while(t--) 93 { 94 scanf("%d",&n); 95 f.clear(); 96 int g = 0; 97 for(i = 1; i <=n ;i++) 98 { 99 double x1,x2,y1,y2; 100 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 101 p[++g] = node(x1,x2,y1,1); 102 que[g] = x1; 103 p[++g] = node(x1,x2,y2,-1); 104 que[g] = x2; 105 } 106 sort(p+1,p+g+1); 107 sort(que+1,que+g+1); 108 int o = 0; 109 f[que[1]] = ++o; 110 val[o] = que[1]; 111 for(i = 2; i <= g; i++) 112 { 113 if(fabs(que[i]-que[i-1])>0) 114 { 115 f[que[i]] = ++o; 116 val[o] = que[i]; 117 } 118 } 119 double ans = 0; 120 build(1,o-1,1); 121 for(i = 1; i < g ; i++) 122 { 123 int l = f[p[i].x1]; 124 int r = f[p[i].x2]-1; 125 if(l<=r) 126 { 127 update(l,r,p[i].f,1,o-1,1); 128 } 129 double sum = s[1]*(p[i+1].y-p[i].y); 130 // cout<<sum<<endl; 131 ans+=sum; 132 } 133 printf("%.2f ",ans); 134 } 135 return 0; 136 }