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
题意：
给你n个机器人的坐标，并且告诉你每一个机器人可以走的方向（只有4个方向），找你找出一个所有机器人都能到达的点。
思路：
维护四个值，最右的x，最左的x，最上的y，最下的y。刚开始定为最大值，构成一个全局矩阵。
根据机器人的坐标和可行方向来缩小能走到的范围，最后判断还能否构造矩阵，能就输出矩阵的任意一个数，否则输出0
本地比赛时写复杂了，多维护了很多无用的变量，并且上下左右没有根据x+-1来确定清楚，导致浪费时间，反思一下。
细节见代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
const int MAXC = 1e5;
int main() {
int q;
cin >> q;
while (q--) {
int n;
cin >> n;
int mnx = -MAXC, mxx = MAXC;
int mny = -MAXC, mxy = MAXC;
while (n--) {
int x, y, f1, f2, f3, f4;
cin >> x >> y >> f1 >> f2 >> f3 >> f4;
if (!f1) mnx = max(mnx, x);
if (!f2) mxy = min(mxy, y);
if (!f3) mxx = min(mxx, x);
if (!f4) mny = max(mny, y);
}
if (mnx <= mxx && mny <= mxy)
cout << "1 " << mnx << " " << mny << "\n";
else
cout << "0\n";
}
return 0;
}原文地址：https://www.cnblogs.com/qieqiemin/p/11247861.html