POJ1177 HDU1828 Picture，线段树求矩形并周长

和矩形并求面积类似，不过T_NODE中保存num、lb、rb用于记录段信息，num记录的是这个节点共有多少个不连续的段，lb、rb记录的是这个节点的上下界是否与别的节点有重合。

/******************************************************************************* # Author : Neo Fung # Email : [e-mail] # Last modified: 2012-01-25 18:57 # Filename: POJ1177 HDU1828 Picture.cpp # Description : 线段树求矩形并 ******************************************************************************/#ifdef _MSC_VER#define DEBUG#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MAX 5010using namespace std;//length当前节点子树覆盖的长度//num，当前节点子树中被分为多少段不连续的小段，用于计算平行于X轴的线段长度//cover，标记当前节点是否被覆盖//lb，rb，标记当前节点的左端点和右端点是否被覆盖，找num的数量需要用到。struct T_NODE{int l,r,length,cover,num;bool lb,rb;}node[MAX<<2];struct LINE{int x,y1,y2,flag;}line[MAX<<1];int y_coord[MAX<<1];void add_line(const int &x1,const int &x2,const int &y1,const int &y2,int &cnt){line[cnt].x=x1;line[cnt].y1=y1; line[cnt].y2=y2; line[cnt].flag=1;y_coord[cnt++]=y1;line[cnt].x=x2;line[cnt].y1=y1; line[cnt].y2=y2; line[cnt].flag=-1;y_coord[cnt++]=y2;}bool inline cmp(const LINE &lhs,const LINE &rhs){return lhs.x<rhs.x;}void init(){memset(node,0,sizeof(node));}void build(const int &t,const int &l,const int &r){node[t].l=l;node[t].r=r;node[t].num = 0;if(l==r-1) return;int mid=(l+r)>>1;build(L(t),l,mid);build(R(t),mid,r);}void update_len(const int &t){if(node[t].cover>0){node[t].num = node[t].lb =node[t].rb =1;node[t].length = y_coord[node[t].r]-y_coord[node[t].l];}else if(node[t].l==node[t].r-1)node[t].length=node[t].num=node[t].lb=node[t].rb=0;else{node[t].length = node[R(t)].length+node[L(t)].length;node[t].rb=node[R(t)].rb;node[t].lb=node[L(t)].lb;//如果两段的端点重合，则-1node[t].num=node[L(t)].num+node[R(t)].num - node[L(t)].rb*node[R(t)].lb;}}void update_tree(const int &t,const LINE &lne){if(y_coord[node[t].l]>=lne.y1 && y_coord[node[t].r]<=lne.y2){node[t].cover+=lne.flag;update_len(t);return;}if(node[t].l==node[t].r-1)return;int mid = (node[t].l+node[t].r)>>1;if(lne.y1<=y_coord[mid])update_tree(L(t),lne);if(lne.y2>y_coord[mid])update_tree(R(t),lne);update_len(t);}__int64 solve(const int &n,const int &cnt){init();build(1,0,cnt-1);__int64 sum=0ll;  int last =0 ,lines=0;for(int i=0;i<n;++i){update_tree(1,line[i]);if(i>0)sum+=2ll * lines *(line[i].x - line[i-1].x);  //求平行于X轴的边的长度sum += abs(node[1].length - last);  //求平行于Y轴的边的长度last = node[1].length;lines = node[1].num;}return sum;}int main(void){#ifdef DEBUG    freopen("../stdin.txt","r",stdin);  freopen("../stdout.txt","w",stdout); #endif    int n;  int x1,y1,x2,y2;  while(~scanf("%d",&n) && n)  {  int cnt=0;  for(int i=0;i<n;++i)  {  scanf("%d%d%d%d",&x1,&y1,&x2,&y2);  add_line(x1,x2,y1,y2,cnt);  }  sort(line,line+cnt,cmp);  sort(y_coord,y_coord+cnt);  int temp=unique(y_coord,y_coord+cnt)-y_coord;  printf("%I64d\n",solve(cnt,temp));  }  return 0;}