标签:post long exp case ring data xpl memset class
非常裸的线段树求面积并。
坐标须要离散化一下。
#include<stdio.h> #include<iostream> #include<stdlib.h> #include<string.h> #include<algorithm> #include<vector> #include<math.h> #include<map> #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define maxn 11000 #define mem(a,b) (memset(a),b,sizeof(a)) #define lmin 1 #define rmax len #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define root lmin,rmax,1 #define now l,r,rt #define int_now int l,int r,int rt #define INF 99999999 #define LL long long #define mod 10007 #define eps 1e-6 #define zero(x) (fabs(x)<eps?0:x) map<double,int>mp; double du[4010]; int len; struct list { double x1,y1; double x2,y2; }node[maxn]; struct linen { double y1,y2; int leap; double x; friend bool operator <(const linen &a,const linen &b) { if(zero(a.x-b.x)!=0) return a.x<b.x; else return a.leap>b.leap; } }line[maxn*2]; double num[maxn*4*4*2]; int cover[maxn*4*4*2]; void push_up(int_now) { if(cover[rt]==0) { num[rt]=num[rt<<1]+num[rt<<1|1]; } if(cover[rt]>=1) { num[rt]=du[r+1]-du[l]; } // cout<<rt<<" "<<du[l]<<"===="<<du[r+1]<<" "<<cover[rt]<<" "<<num[rt]<<" "<<sum[rt]<<endl; } void push_down(int_now) { } void creat(int_now) { memset(cover,0,sizeof(cover)); memset(num,0,sizeof(num)); } void updata(int ll,int rr,int x,int_now) { if(ll>r||rr<l)return; if(ll<=l&&rr>=r) { // cout<<l<<" "<<r<<"-"<<du[l]<<" "<<du[r+1]<<" "<<endl; cover[rt]+=x; push_up(now); return; } updata(ll,rr,x,lson); updata(ll,rr,x,rson); push_up(now); } int main() { int t,n,s; int cas=0; while(~scanf("%d",&n)&&n) { cas++; s=0; for(int i=1;i<=n;i++) { scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2); du[++s]=node[i].x1; du[++s]=node[i].y1; du[++s]=node[i].x2; du[++s]=node[i].y2; line[i*2-1].x=node[i].x1; line[i*2-1].y1=node[i].y1; line[i*2-1].y2=node[i].y2; line[i*2-1].leap=1; line[i*2].x=node[i].x2; line[i*2].y1=node[i].y1; line[i*2].y2=node[i].y2; line[i*2].leap=-1; } sort(line+1,line+n*2+1); sort(du+1,du+s+1); du[0]=-1; len =0; mp.clear(); for(int i=1;i<=s;i++) { if(du[i]!=du[i-1]) { mp[du[i]]=++len; du[len]=du[i]; } } creat(root); double st=0; double are=0.0; for(int i=1;i<=len;i++) { // cout<<i<<"----"<<du[i]<<endl; } len--; // cout<<1<<" "<<len<<" "<<1<<" "<<endl; for(int i=1;i<=n*2;i++) { int l,r; l=mp[line[i].y1]; r=mp[line[i].y2]; if(zero(line[i].x-st)!=0) { are+=(line[i].x-st)*num[1]; st=line[i].x; } // cout<<line[i].x<<" "<<line[i].y1<<" "<<line[i].y2<<" "<<" "<<line[i].leap<<endl; updata(l,r-1,line[i].leap,root); //cout<<l<<" "<<r<<" "<<are<<endl; } printf("Test case #%d\n",cas); printf("Total explored area: %.2f\n\n",are); } return 0; }
标签:post long exp case ring data xpl memset class
原文地址:http://www.cnblogs.com/brucemengbm/p/6923499.html