[POJ1151]Atlantis
试题描述
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.
输入
The input 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.
输出
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.
输入示例
2 10 10 20 20 15 15 25 25.5 0
输出示例
Test case #1 Total explored area: 180.00
数据规模及约定
见“输入”
题解
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 4010
int n, cnt, ca, cd;
struct Rec {
double x1, y1, x2, y2;
Rec() {}
Rec(double _1, double _2, double _3, double _4): x1(_1), y1(_2), x2(_3), y2(_4) {}
} rs[maxn];
struct Rec_int { int x1, y1, x2, y2; } rsi[maxn];
double num[maxn<<1], A[maxn<<1], B[maxn<<1];
struct Line {
int l, r, x;
Line() {}
Line(int _1, int _2, int _3): l(_1), r(_2), x(_3) {}
bool operator < (const Line& t) const { return x < t.x; }
} ad[maxn], de[maxn];
double sumv[maxn<<3];
int addv[maxn<<3];
void build(int L, int R, int o) {
if(L == R) {
sumv[o] = 0.0;
addv[o] = 0;
return ;
}
int M = L + R >> 1, lc = o << 1, rc = lc | 1;
build(L, M, lc); build(M+1, R, rc);
sumv[o] = 0.0; addv[o] = 0;
return ;
}
void update(int L, int R, int o, int ql, int qr, int v) {
int M = L + R >> 1, lc = o << 1, rc = lc | 1;
if(ql <= L && R <= qr) {
addv[o] += v;
if(addv[o]) sumv[o] = A[R] - A[L-1];
else if(L == R) sumv[o] = 0.0;
else sumv[o] = sumv[lc] + sumv[rc];
return ;
}
if(ql <= M) update(L, M, lc, ql, qr, v);
if(qr > M) update(M+1, R, rc, ql, qr, v);
sumv[o] = addv[o] ? A[R] - A[L-1] : sumv[lc] + sumv[rc];
return ;
}
int main() {
int kase = 0;
while(1) {
scanf("%d", &n);
if(!n) break;
cnt = 0;
for(int i = 1; i <= n; i++) {
double x1, x2, y1, y2;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
rs[i] = Rec(x1, y1, x2, y2);
num[++cnt] = x1; num[++cnt] = x2;
}
sort(num + 1, num + cnt + 1);
cnt = unique(num + 1, num + cnt + 1) - num - 1;
int tcnt = cnt;
for(int i = 1; i < cnt; i++) A[i] = num[i+1] - num[1];
for(int i = 1; i <= n; i++) {
rsi[i].x1 = lower_bound(num + 1, num + cnt + 1, rs[i].x1) - num;
rsi[i].x2 = lower_bound(num + 1, num + cnt + 1, rs[i].x2) - num - 1;
// printf("%d %d
", rsi[i].x1, rsi[i].x2);
}
cnt = 0;
for(int i = 1; i <= n; i++)
num[++cnt] = rs[i].y1, num[++cnt] = rs[i].y2;
sort(num + 1, num + cnt + 1);
cnt = unique(num + 1, num + cnt + 1) - num - 1;
for(int i = 1; i < cnt; i++) B[i] = num[i+1] - num[i];
ca = cd = 0;
for(int i = 1; i <= n; i++) {
rsi[i].y1 = lower_bound(num + 1, num + cnt + 1, rs[i].y1) - num;
rsi[i].y2 = lower_bound(num + 1, num + cnt + 1, rs[i].y2) - num;
ad[++ca] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y1);
de[++cd] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y2);
}
sort(ad + 1, ad + ca + 1);
// for(int i = 1; i <= ca; i++) printf("[%d %d] %d
", ad[i].l, ad[i].r, ad[i].x);
sort(de + 1, de + cd + 1);
double ans = 0.0;
int ka = 1, kd = 1;
build(1, tcnt, 1);
for(int i = 1; i <= cnt; i++) {
while(ka <= ca && ad[ka].x == i) {
// printf("add: [%d, %d]
", ad[ka].l, ad[ka].r);
update(1, tcnt, 1, ad[ka].l, ad[ka].r, 1), ka++;
}
while(kd <= cd && de[kd].x == i) {
// printf("del: [%d, %d]
", de[kd].l, de[kd].r);
update(1, tcnt, 1, de[kd].l, de[kd].r, -1), kd++;
}
if(i < cnt) ans += sumv[1] * B[i];
// printf("sumv[1]: %.2lf
", sumv[1]);
}
printf("Test case #%d
Total explored area: %.2lf
", ++kase, ans);
}
return 0;
}