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Description
Input
Output
Sample Input
2 10 10 20 20 15 15 25 25.5 0
Sample Output
Test case #1 Total explored area: 180.00
题目就是求所有矩形的并面积。
通过查阅知道了是扫描线,了解了扫描线的原理,用线段树手写了一下,结果PushUp函数写搓了。。看了AC的代码才知道了原因。
做法就是通过对纵坐标有序化,然后创建区间。
然后通过横向扫描过去,得到每段横向段的高度,乘以宽度就是面积了。
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
//线段树
//扫描线
const int maxn = 205;
struct node
{
int lt, rt;
double height;
int num;
}tree[4*maxn];
struct Line
{
double x;
double y1, y2;
bool isLeft;
}line[maxn];
bool cmp(Line a, Line b)
{
return a.x < b.x;
}
double y[maxn];
//向上更新
void PushUp(int id)
{
if(tree[id].num > 0)
{
tree[id].height = y[tree[id].rt] - y[tree[id].lt];
return;
}
if(tree[id].lt+1 == tree[id].rt)
tree[id].height = 0;
else
tree[id].height = tree[id<<1].height + tree[id<<1|1].height;
}
//建立线段树
void Build(int lt, int rt, int id)
{
tree[id].lt = lt;
tree[id].rt = rt;
tree[id].height = 0;//每段的初值,根据题目要求
tree[id].num = 0;
if (lt+1 == rt)
{
//tree[id].val = 1;
return;
}
int mid = (lt + rt) >> 1;
Build(lt, mid, id<<1);
Build(mid, rt, id<<1|1);
//PushUp(id);
}
//寻找符合修改的区间通过判断num进行修改
void Updata(int id,Line p)
{
if(p.y1 <= y[tree[id].lt] && p.y2 >= y[tree[id].rt])
{
if (p.isLeft > 0)
tree[id].num++;
else
tree[id].num--;
PushUp(id);
return;
}
int mid = (tree[id].lt+tree[id].rt) >> 1;
if (p.y1 < y[mid])
Updata(id<<1, p);
if (p.y2 > y[mid])
Updata(id<<1|1, p);
PushUp(id);
}
int n;
void Input()
{
double x1, y1, x2, y2;
int cnt = 1;
for (int i = 0; i < n; ++i)
{
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
y[cnt] = y1;
y[cnt+1] = y2;
line[cnt].x = x1;
line[cnt].y1 = y1;
line[cnt].y2 = y2;
line[cnt].isLeft = true;
line[cnt+1].x = x2;
line[cnt+1].y1 = y1;
line[cnt+1].y2 = y2;
line[cnt+1].isLeft = false;
cnt += 2;
}
sort(y+1, y+1+2*n);
sort(line+1, line+1+2*n, cmp);
Build(1, 2*n, 1);
}
double Work()
{
double ans = 0;
Updata(1, line[1]);
int len = 2*n;
for (int i = 2; i <= len; ++i)
{
ans += (line[i].x-line[i-1].x) * tree[1].height;
Updata(1, line[i]);
}
return ans;
}
int main()
{
//freopen("test.in", "r", stdin);
int times = 1;
while (scanf("%d", &n) != EOF && n)
{
Input();
double ans = Work();
printf("Test case #%d\n", times);
printf("Total explored area: %.2lf\n\n", ans);
times++;
}
return 0;
}
ACM学习历程—POJ1151 Atlantis(扫描线 && 线段树)
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原文地址:http://www.cnblogs.com/andyqsmart/p/4475579.html