73th LeetCode Weekly Contest Escape The Ghosts

You are playing a simplified Pacman game. You start at the point

(0, 0)

, and your destination is

(target[0], target[1])

. There are several ghosts on the map, the i-th ghost starts at

(ghosts[i][0], ghosts[i][1])

.

Each turn, you and all ghosts simultaneously *may* move in one of 4 cardinal directions: north, east, west, or south, going from the previous point to a new point 1 unit of distance away.

You escape if and only if you can reach the target before any ghost reaches you (for any given moves the ghosts may take.) If you reach any square (including the target) at the same time as a ghost, it doesn't count as an escape.

Return True if and only if it is possible to escape.

Example 1:
Input:
ghosts = [[1, 0], [0, 3]]
target = [0, 1]
Output: true
Explanation:
You can directly reach the destination (0, 1) at time 1, while the ghosts located at (1, 0) or (0, 3) have no way to catch up with you.

Example 2:
Input:
ghosts = [[1, 0]]
target = [2, 0]
Output: false
Explanation:
You need to reach the destination (2, 0), but the ghost at (1, 0) lies between you and the destination.

Example 3:
Input:
ghosts = [[2, 0]]
target = [1, 0]
Output: false
Explanation:
The ghost can reach the target at the same time as you.

Note:

All points have coordinates with absolute value <=

10000

.

The number of ghosts will not exceed

100

.

鬼和玩家都可以四周移动，问玩家到达目标地方之前会不会被鬼遇见，在目标点遇见也是不行的哦

当然是看谁离目标近啊！！！！

class Solution {
public:
bool escapeGhosts(vector<vector<int>>& ghosts, vector<int>& target) {
int distar=abs(target[0])+abs(target[1]);
int len=ghosts.size();
for(int i=0;i<len;i++){
for(int j=0;j<2;j++){
int ghostsdis=abs(ghosts[i][0]-target[0])+abs(ghosts[i][1]-target[1]);
if(ghostsdis<=distar){
return false;
}
}
}
return true;
}