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codeforce1196C

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dejavu1zz
发布2020-10-23 15:14:43
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发布2020-10-23 15:14:43
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题意描述

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.

思路

遍历一遍机器人的坐标以及机器人所能进行的操作,求出向左最大能到达的坐标,向右最小能到达的坐标,向上和向下能到达的坐标,最后判断向左和向下能够到达的坐标是否小于向右和向上能够到达的坐标,输出左下的坐标即可。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#pragma GCC optimize(2)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC diagnostic error "-std=c++14"
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef long long LL;
const int N=1e5+10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
int main(){
    IOS;
    int T;cin>>T;
    while(T--){
        int n;cin>>n;
        int maxx=100000,minx=-100000,maxy=100000,miny=-100000;
        for(int i=0;i<n;i++){
            int x,y,l,r,u,d;cin>>x>>y>>l>>u>>r>>d;
            if(!l) minx=max(minx,x);
            if(!u) maxy=min(maxy,y);
            if(!r) maxx=min(maxx,x);
            if(!d) miny=max(miny,y);
        }
        if(minx<=maxx && miny<=maxy){
            cout<<"1 "<<minx<<' '<<miny<<endl;
        }else{
            cout<<"0"<<endl;
        }
    }
    return 0;
}
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原始发表:2020-05-03 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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  • 题意描述
  • 思路
  • AC代码
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