HDU1542 Atlantis —— 求矩形面积并 线段树 + 扫描线 + 离散化
题目链接:https://vjudge.net/problem/HDU-1542
InputThe 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.OutputFor 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.
Sample Input
2
10 10 20 20
15 15 25 25.5
0
Sample Output
Test case #1
Total explored area: 180.00
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e5+; struct line
{
double le, ri, h;
int id;
bool operator<(const line &a)const{
return h<a.h;
}
}Line[MAXN]; //X用于离散化横坐标,times为此区间被覆盖的次数,sum为此区间被覆盖的长度
double X[MAXN], times[MAXN<<], sum[MAXN<<]; void push_up(int u, int l, int r)
{
if(times[u]>) //该区间被覆盖,则覆盖长度为区间长度
sum[u] = X[r] - X[l];
else //该区间没有被覆盖,如果为单位区间,则覆盖长度为0,否则为两个子区间的覆盖长度之和。
sum[u] = (l+==r)?:sum[u*]+sum[u*+];
} //此种线段树的操作对象为连续型,即最小的元素为长度为1的区间[l,r],其中l和r只代表端点(r-l>=1),用于确定
//区间的位置和长度,l和r本身没有特别的含义。而以往做的什么单点更新之类的,都属于离散型,在l处和r处是有含义的
void add(int u, int l, int r, int x, int y, int v)
{
if(x<=l && r<=y)
{
times[u] += v;
push_up(u, l, r);
return;
} int mid = (l+r)>>;
if(x<=mid-) add(u*, l, mid, x, y, v);
if(y>=mid+) add(u*+, mid, r, x, y, v);
push_up(u, l, r);
} int main()
{
int n, kase = ;
while(scanf("%d", &n) && n)
{
for(int i = ; i<=n; i++)
{
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
Line[i].le = Line[i+n].le = x1;
Line[i].ri = Line[i+n].ri = x2;
Line[i].h = y1; Line[i+n].h = y2;
Line[i].id = ; Line[i+n].id = -;
X[i] = x1; X[i+n] = x2;
} sort(Line+, Line++*n);
sort(X+, X++*n);
int m = unique(X+, X++*n) - (X+); //去重 memset(sum, , sizeof(sum));
memset(times, , sizeof(times)); double ans = ;
for(int i = ; i<=*n-; i++)
{
int l = upper_bound(X+, X++m, Line[i].le) - (X+);
int r = upper_bound(X+, X++m, Line[i].ri) - (X+);
add(, , m, l, r, Line[i].id);
ans += sum[]* (Line[i+].h-Line[i].h);
}
printf("Test case #%d\n", ++kase);
printf("Total explored area: %.2f\n\n", ans);
}
}
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