hdu 1542 线段树
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Atlantis
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12402 Accepted Submission(s): 5203
Problem Description
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the
total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
Input
The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000),
not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don’t process it.
The input file is terminated by a line containing a single 0. Don’t process it.
Output
For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area
(i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Output a blank line after each test case.
Sample Input
2 10 10 20 20 15 15 25 25.5 0
Sample Output
Test case #1 Total explored area: 180.00
Source
Recommend
模拟：
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=110;
struct LINE
{
double x, y_down, y_up;
int flag;
bool operator<(const LINE &a)const
{
return x<a.x;
}
}line[2*maxn];
struct TREE
{
double y_down, y_up;
double x;
int cover; //用以表示加进线段树中的线段次数
bool flag; //此标记用来表示是否有超元线段；为了处理方便加上去的
}tree[1000*maxn];
int n;
double x1, y1, x2, y2;
int index=0;
double y[2*maxn];
void build(int i, int l, int r)
{
tree[i].x = -1; //-1表示该区间已经没有线段
tree[i].cover = 0; //表示该区间上有多少条线段；左边线段加进去则++，右边线段加进去则--
tree[i].y_down = y[l];
tree[i].y_up = y[r];
tree[i].flag = false;
if(l+1==r)
{
tree[i].flag = true; //flag==true表示达到了叶子节点
return;
}
int mid=(l+r)>>1;
build(2*i, l, mid);
build(2*i+1, mid, r);
}
double insert(int i, double x, double l, double r, int flag) //flag表示为左边还是右边
{
if (r<=tree[i].y_down || l>=tree[i].y_up)
return 0;
if (tree[i].flag)
{
if (tree[i].cover > 0) //递归到了叶子节点
{
double temp_x = tree[i].x;
double ans=(x-temp_x)*(tree[i].y_up - tree[i].y_down);
tree[i].x = x; //定位上一次的x
tree[i].cover += flag;
return ans;
}
else
{
tree[i].cover += flag;
tree[i].x = x;
return 0;
}
}
double ans1, ans2;
ans1 = insert(2*i, x, l, r, flag);
ans2 = insert(2*i+1, x, l, r, flag);
return ans1+ans2;
}
int main( )
{
// freopen("d:\\in.txt","r",stdin);
int count=0;
while (scanf("%d", &n)!=EOF&&n)
{
index = 1;
for (int i=1; i<=n; i++)
{
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
y[index] = y1;
line[index].x = x1;
line[index].y_down = y1;
line[index].y_up = y2;
line[index].flag = 1; //1表示左边
index++;
y[index] = y2;
line[index].x = x2;
line[index].y_down = y1;
line[index].y_up = y2;
line[index].flag = -1; //-1表示右边
index++;
}
sort(&y[1], &y[index]); //把所有的纵坐标按从小到大排序,把1写成了0，WA一次
sort(&line[1], &line[index]);
build(1, 1, index-1);
double ans=0;
for (i=1;i<index; i++)
{
ans += insert(1, line[i].x, line[i].y_down, line[i].y_up, line[i].flag);
}
printf("Test case #%d\nTotal explored area: %.2f\n\n", ++count, ans);
}
return 0;
}
树形DP#include<iostream>
#include<string>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct node
{
int l;
int r;
int cover;
double len;
};
node tree[2000];
double yy[250];
int n,len;
struct Line
{
double y_down;
double y_up;
double x;
int cover;
};
Line line[250];
int cmp(Line a,Line b)//根据线段的x左边排序
{
return a.x<b.x;
}
int find(double x)
{
int l=0,r=len,mid;
while(l<=r)
{
mid=(l+r)/2;
if(yy[mid]==x)
return mid;
if(yy[mid]<x)
l=mid+1;
else
r=mid-1;
}
return l;
}
void build(int i,int l,int r)//建树，按照yy数组的下标建搜索树
{
tree[i].l=l;
tree[i].r=r;
tree[i].cover=0;
tree[i].len=0;
if(l+1==r)
return;
int mid=(l+r)/2;
build(2*i,l,mid);
build(2*i+1,mid,r);
}
void fun(int i)
{
if(tree[i].cover)
tree[i].len=yy[tree[i].r]-yy[tree[i].l]; //如果cover大于1，那么整段都可用于与下一线段求并面积
else if(tree[i].l+1==tree[i].r) //叶子线段
tree[i].len=0;
else
tree[i].len=tree[2*i].len+tree[2*i+1].len; //很简单的dp
}
void updata(int i,int l,int r,int cover)
{
if(tree[i].l>r || tree[i].r<l)
return;
if(tree[i].l>=l && tree[i].r<=r)
{
tree[i].cover+=cover;
fun(i);
return;
}
updata(2*i,l,r,cover);
updata(2*i+1,l,r,cover);
fun(i);
}
int main()
{
double x1,y1,x2,y2;
int i,m,a,b,cas=1;
while(scanf("%d",&n)==1 && n)
{
m=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
yy[m]=y1;
line[m].cover=1;
line[m].x=x1;
line[m].y_down=y1;
line[m++].y_up=y2;
yy[m]=y2;
line[m].cover=-1;
line[m].x=x2;
line[m].y_down=y1;
line[m++].y_up=y2;
}
sort(yy,yy+m);//给横坐标的点排序
len=1;
for(i=1;i<m;i++)//去重复的线段
{
if(yy[i-1]!=yy[i])
yy[len++]=yy[i];
}
len--;
sort(line,line+m,cmp);
build(1,0,len);
double ans=0;
printf("Test case #%d\n",cas++);
for(i=0;i<m-1;i++)
{
a=find(line[i].y_down);
b=find(line[i].y_up);
updata(1,a,b,line[i].cover);
ans+=tree[1].len*(line[i+1].x-line[i].x); //tree[1].len是侧边的长，line[i+1].x-line[i].x=底边长
}
printf("Total explored area: %0.2lf\n\n",ans);
}
return 0;
}
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