【BZOJ】【2178】圆的面积并
自适应辛普森积分
Orz Hzwer
辛普森真是个强大的东西……很多东西都能积= =
这题的正解看上去很鬼畜，至少我这种不会计算几何的渣渣是写不出来……（对圆的交点求图包，ans=凸包的面积+一堆弓形的面积，另外还有中空的情况……那种凸包怎么求啊喂！）
1 /************************************************************** 2 Problem: 2178 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:8808 ms 7 Memory:1372 kb 8 ****************************************************************/ 9 10 //BZOJ 2178 11 #include<cmath> 12 #include<vector> 13 #include<cstdio> 14 #include<cstring> 15 #include<cstdlib> 16 #include<iostream> 17 #include<algorithm> 18 #define rep(i,n) for(int i=0;i<n;++i) 19 #define F(i,j,n) for(int i=j;i<=n;++i) 20 #define D(i,j,n) for(int i=j;i>=n;--i) 21 #define pb push_back 22 using namespace std; 23 inline int getint(){ 24 int v=0,sign=1; char ch=getchar(); 25 while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();} 26 while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();} 27 return v*sign; 28 } 29 const int N=1010,INF=~0u>>2; 30 const double eps=1e-13; 31 typedef long long LL; 32 typedef double db; 33 /******************tamplate*********************/ 34 int n,top,st,ed; 35 db xl[N],xr[N],ans; 36 bool del[N]; 37 struct data{db x,y,r;}t[N],sk[N]; 38 struct line{db l,r;}p[N]; 39 db dis(data a,data b){ 40 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 41 } 42 bool cmp1(data a,data b){return a.r<b.r;} 43 bool cmp2(data a,data b){return a.x-a.r<b.x-b.r;} 44 bool cmp3(line a,line b){return a.l<b.l;} 45 void init(){ 46 scanf("%d",&n); 47 F(i,1,n) scanf("%lf%lf%lf",&t[i].x,&t[i].y,&t[i].r); 48 sort(t+1,t+n+1,cmp1); 49 F(i,1,n) F(j,i+1,n) 50 if(dis(t[i],t[j])<=t[j].r-t[i].r){ 51 del[i]=1; break; 52 } 53 F(i,1,n)if(!del[i]) sk[++top]=t[i];n=top; 54 sort(sk+1,sk+n+1,cmp2); 55 } 56 db getf(db x){ 57 int j,sz=0; db r,len=0,dis; 58 F(i,st,ed){ 59 if (x<=xl[i]||x>=xr[i])continue; 60 dis=sqrt(sk[i].r-(x-sk[i].x)*(x-sk[i].x)); 61 p[++sz].l=sk[i].y-dis; p[sz].r=sk[i].y+dis; 62 } 63 sort(p+1,p+sz+1,cmp3); 64 F(i,1,sz){ 65 r=p[i].r; 66 for(j=i+1;j<=sz;j++){ 67 if(p[j].l>r)break; 68 if (r<p[j].r) r=p[j].r; 69 } 70 len+=r-p[i].l; i=j-1; 71 } 72 return len; 73 } 74 db cal(db l,db fl,db fmid,db fr){ 75 return (fl+fmid*4+fr)*l/6; 76 } 77 db simpson(db l,db mid,db r,db fl,db fmid,db fr,db s){ 78 db m1=(l+mid)/2,m2=(mid+r)/2; 79 db f1=getf(m1),f2=getf(m2); 80 db g1=cal(mid-l,fl,f1,fmid),g2=cal(r-mid,fmid,f2,fr); 81 if (fabs(g1+g2-s)<eps)return g1+g2; 82 return simpson(l,m1,mid,fl,f1,fmid,g1)+ 83 simpson(mid,m2,r,fmid,f2,fr,g2); 84 } 85 void work(){ 86 int i,j; 87 db l,r,mid,fl,fr,fmid; 88 F(i,1,n){ 89 xl[i]=sk[i].x-sk[i].r; 90 xr[i]=sk[i].x+sk[i].r; 91 sk[i].r*=sk[i].r; 92 } 93 F(i,1,n){ 94 l=xl[i]; r=xr[i]; 95 for(j=i+1;j<=n;j++){ 96 if (xl[j]>r) break; 97 if (xr[j]>r) r=xr[j]; 98 } 99 st=i; ed=j-1;i=j-1; 100 mid=(l+r)/2; 101 fl=getf(l); fr=getf(r); fmid=getf(mid); 102 ans+=simpson(l,mid,r,fl,fmid,fr,cal(r-l,fl,fmid,fr)); 103 } 104 } 105 int main(){ 106 #ifndef ONLINE_JUDGE 107 freopen("2178.in","r",stdin); 108 freopen("2178.out","w",stdout); 109 #endif 110 init(); 111 work(); 112 printf("%.3lf",ans); 113 return 0; 114 }
2178: 圆的面积并
Time Limit: 20 Sec Memory Limit: 259 MBSubmit: 908 Solved: 218
[Submit][Status][Discuss]
Description
给出N个圆，求其面积并
Input
先给一个数字N ,N< = 1000
接下来是N行是圆的圆心，半径，其绝对值均为小于1000的整数
Output
面积并，保留三位小数
Sample Input
721
。。。。。。
Sample Output
12707279.690
