POJ 1151 线段树 扫描线 模板题

第一道扫描线求相交矩阵面积的题 

矩阵面积并的模板题  同HDU 1542

建树的方法和原来的线段树不太一样

#include "stdio.h"#include "string.h"#include "stdlib.h"#include "math.h"#include "iostream"#include "algorithm"using namespace std;struct comp{int flag; // 1代表矩阵的起始边，-1代表矩阵的终边double x,y1,y2;} node[300];  // 记录每条与Y轴平行的边struct comp1{int l,r,mid,s;double len,ml,mr;} data[1001]; bool cmp(comp a,comp b){return a.x-b.x<0.0000001;}double y[300];void build(int l,int r,int k){data[k].l=l;data[k].r=r;data[k].mid=(l+r)/2;data[k].ml=y[l];data[k].mr=y[r];data[k].s=data[k].len=0;if (l+1==r) return ;build(l,data[k].mid,k*2);build(data[k].mid,r,k*2+1); // 注意应该从 mid开始}void callen(int k) // 计算纵长度{if (data[k].s>0) data[k].len=data[k].mr-data[k].ml;else if (data[k].l+1==data[k].r) data[k].len=0;else data[k].len=data[k*2].len+data[k*2+1].len;}void updata(int k,comp b){if (b.y1==data[k].ml && b.y2==data[k].mr) {data[k].s+=b.flag;callen(k);return ;}if (b.y2<=data[k*2].mr) updata(k*2,b);else if (b.y1>=data[k*2+1].ml) updata(k*2+1,b);else {comp temp;temp=b;temp.y2=data[k*2].mr;updata(k*2,temp);temp=b;temp.y1=data[k*2+1].ml;updata(k*2+1,temp);}callen(k);}int main(){int n,i,t,p;double sum,x1,x2,y1,y2;p=1;while (scanf("%d",&n),n!=0){t=1;for (i=0;i<n;i++){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);node[t].x=x1;node[t].y1=y1;node[t].y2=y2;node[t].flag=1;y[t++]=y1;node[t].x=x2;node[t].y1=y1;node[t].y2=y2;node[t].flag=-1;y[t++]=y2;}sort(node+1,node+t,cmp);sort(y+1,y+t);build(1,t-1,1);sum=0;updata(1,node[1]);for (i=2;i<t;i++){sum+=data[1].len*(node[i].x-node[i-1].x);updata(1,node[i]);}printf("Test case #%d\n",p++);printf("Total explored area: %.2lf\n\n",sum);}return 0;}