2019 Hang electric multi-school 81009
Topic links: HDU 6665
Game link: 2019 Multi-University Training Contest The 8
Problem Description
Calabash is the servant of a landlord. The landlord owns a piece of land, which can be regarded as an infinite 2D plane.
One day the landlord set up two orthogonal rectangular-shaped fences on his land. He asked Calabash a simple problem: how many nonempty connected components is my land divided into by these two fences, both finite and infinite? Calabash couldn't answer this simple question. Please help him!
Recall that a connected component is a maximal set of points not occupied by the fences, and every two points in the set are reachable without crossing the fence.
Input
The first line of input consists of a single integer \(T (1\le T\le 10000)\), the number of test cases.
Each test case contains two lines, specifying the two rectangles. Each line contains four integers \(x_1,y_1,x_2,y_2 (0\le x_1,y_1,x_2,y_2\le 10^9,x_1<x_2,y_1<y_2)\), where \((x_1,y_1),(x_2,y_2)\) are the Cartesian coordinates of two opposite vertices of the rectangular fence. The edges of the rectangles are parallel to the coordinate axes. The edges of the two rectangles may intersect, overlap, or even coincide.
Output
For each test case, print the answer as an integer in one line.
Sample Input
3
0 0 1 1
2 2 3 4
1 0 3 2
0 1 2 3
0 0 1 1
0 0 1 1Sample Output
3
4
2Solution
Meaning of the questions:
Given two rectangles, seeking the plane into rectangular pieces.
Thinking
Category talk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
struct Point
{
ll x, y;
}p[10];
int main() {
int T;
cin >> T;
while (T--) {
ll maxx = 0, maxy = 0, minx = inf, miny = inf;
for(int i = 1; i <= 2; ++i) {
scanf("%lld%lld", &p[i].x, &p[i].y);
}
p[3].x = p[1].x;
p[3].y = p[2].y;
p[4].x = p[2].x;
p[4].y = p[1].y;
ll s1 = (p[2].x - p[1].x) * (p[2].y - p[1].y);
for(int i = 5; i <= 6; ++i) {
scanf("%lld%lld", &p[i].x, &p[i].y);
}
p[7].x = p[5].x;
p[7].y = p[6].y;
p[8].x = p[6].x;
p[8].y = p[5].y;
ll s2 = (p[6].x - p[5].x) * (p[6].y - p[5].y);
for(int i = 1; i <= 8; ++i) {
maxx = max(maxx, p[i].x);
maxy = max(maxy, p[i].y);
minx = min(minx, p[i].x);
miny = min(miny, p[i].y);
}
// for(int i = 1; i <= 8; ++i) {
// cout << p[i].x << " " << p[i].y << endl;
// }
if(p[1].x == p[5].x && p[1].y == p[5].y && p[3].x == p[7].x && p[3].y == p[7].y && p[6].x == p[2].x && p[6].y == p[2].y && p[4].x == p[8].x && p[4].y == p[8].y) {
printf("2\n");
continue;
}
ll s = (maxx - minx) * (maxy - miny); // cout << s << endl;
if(s1 == s || s2 == s) {
if((p[1].x == p[5].x && p[2].x == p[6].x)) {
if(p[5].y == p[1].y || p[6].y == p[2].y) printf("3\n");
else printf("4\n");
} else if((p[1].y == p[5].y && p[2].y == p[6].y)) {
if(p[5].x == p[1].x || p[6].x == p[2].x) printf("3\n");
else printf("4\n");
}
else printf("3\n");
} else if(p[4].y >= p[7].y || p[8].y >= p[3].y || p[5].x >= p[2].x || p[1].x >= p[6].x) {
printf("3\n");
} else if((p[2].x - p[1].x) * (p[6].y - p[5].y) == s) {
if(p[6].y > p[2].y && p[5].y < p[1].y && p[1].x < p[5].x && p[2].x > p[6].x) {
printf("6\n");
} else if(p[1].x == p[5].x && p[1].y == p[5].y) {
printf("4\n");
} else if(p[3].x == p[7].x && p[3].y == p[7].y) {
printf("4\n");
} else if(p[2].x == p[6].x && p[2].y == p[6].y) {
printf("4\n");
} else if(p[4].x == p[8].x && p[4].y == p[8].y) {
printf("4\n");
} else {
printf("5\n");
}
} else if((p[6].x - p[5].x) * (p[2].y - p[1].y) == s) {
if(p[2].y > p[6].y && p[1].y < p[5].y && p[5].x < p[1].x && p[6].x > p[2].x) {
printf("6\n");
} else if(p[1].x == p[5].x && p[1].y == p[5].y) {
printf("4\n");
} else if(p[3].x == p[7].x && p[3].y == p[7].y) {
// cout << 1 << endl;
printf("4\n");
} else if(p[2].x == p[6].x && p[2].y == p[6].y) {
printf("4\n");
} else if(p[4].x == p[8].x && p[4].y == p[8].y) {
printf("4\n");
} else {
printf("5\n");
}
} else {
printf("4\n");
}
}
return 0;
}