C. Robot Breakout
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
n robots have escaped from your laboratory! You have to find them as soon as possible, because these robots are experimental, and their behavior is not tested yet, so they may be really dangerous!
Fortunately, even though your robots have escaped, you still have some control over them. First of all, you know the location of each robot: the world you live in can be modeled as an infinite coordinate plane, and the i-th robot is currently located at the point having coordinates (xi, yi). Furthermore, you may send exactly one command to all of the robots. The command should contain two integer numbers X and Y, and when each robot receives this command, it starts moving towards the point having coordinates (X, Y). The robot stops its movement in two cases:
either it reaches (X, Y);
or it cannot get any closer to (X, Y).
Normally, all robots should be able to get from any point of the coordinate plane to any other point. Each robot usually can perform four actions to move. Let’s denote the current coordinates of the robot as (xc, yc). Then the movement system allows it to move to any of the four adjacent points:
the first action allows it to move from (xc, yc) to (xc−1, yc);
the second action allows it to move from (xc, yc) to (xc, yc+1);
the third action allows it to move from (xc, yc) to (xc+1, yc);
the fourth action allows it to move from (xc, yc) to (xc, yc−1).
Unfortunately, it seems that some movement systems of some robots are malfunctioning. For each robot you know which actions it can perform, and which it cannot perform.
You want to send a command so all robots gather at the same point. To do so, you have to choose a pair of integer numbers X and Y so that each robot can reach the point (X, Y). Is it possible to find such a point?
Input
The first line contains one integer q (1≤q≤105) — the number of queries.
Then q queries follow. Each query begins with one line containing one integer n (1≤n≤105) — the number of robots in the query. Then n lines follow, the i-th of these lines describes the i-th robot in the current query: it contains six integer numbers xi, yi, fi,1, fi,2, fi,3 and fi,4 (−105≤xi,yi≤105, 0≤fi,j≤1). The first two numbers describe the initial location of the i-th robot, and the following four numbers describe which actions the i-th robot can use to move (fi,j=1 if the i-th robot can use the j-th action, and fi,j=0 if it cannot use the j-th action).
It is guaranteed that the total number of robots over all queries does not exceed 105.
Output
You should answer each query independently, in the order these queries appear in the input.
To answer a query, you should do one of the following:
if it is impossible to find a point that is reachable by all n robots, print one number 0 on a separate line;
if it is possible to find a point that is reachable by all n robots, print three space-separated integers on the same line: 1 X Y, where X and Y are the coordinates of the point reachable by all n robots. Both X and Y should not exceed 105 by absolute value; it is guaranteed that if there exists at least one point reachable by all robots, then at least one of such points has both coordinates not exceeding 105 by absolute value.
Example
inputCopy
4
2
-1 -2 0 0 0 0
-1 -2 0 0 0 0
3
1 5 1 1 1 1
2 5 0 1 0 1
3 5 1 0 0 0
2
1337 1337 0 1 1 1
1336 1337 1 1 0 1
1
3 5 1 1 1 1
outputCopy
1 -1 -2
1 2 5
0
1 -100000 -100000
题目大意：给出点集和点的活动方向，求点集是否能归并到一个点。
乍一看，很难判断如何模拟点集归并到哪一个点。但每个点的行径方向确定了点集的活动区域。
很巧妙的我们只需要模拟点集活动区域的x正半轴，x负半轴，y正半轴，y负半轴的活动端点即可。
#include <cstdio>
int main()
{
int q, n, x, y, f1, f2, f3, f4;
scanf("%d", &q);
for (int i = 0; i < q; ++i)
{
scanf("%d", &n);
int fx = int(1e5);//最右端100000
int fy = fx;//最上端100000
int f_x = -fx;//最左端-100000
int f_y = -fy;//最下端-100000
//模拟活动区域
for (int j = 0; j < n; ++j)
{
scanf("%d%d%d%d%d%d", &x, &y, &f1, &f2, &f3, &f4);
//如果在该方向上不能活动，更新该方向的端值
if (!f3 && fx > x)
fx = x;
if (!f4 && f_y < y)
f_y = y;
if (!f1 && x > f_x)
f_x = x;
if (!f2 && fy > y)
fy = y;
}
if (fx >= f_x && fy >= f_y) //(x, y)点集的活动区域 f_x <= x <= fx f_y <= y <= fy
{
printf("%d %d %d\n", 1, fx, fy);
}
else
{
printf("0\n");
}
}
return 0;
}
