javascript - 将自行车分配给其他人-第一优先(最接近的自行车最接近的人)

通过网格将功能传递给自行车和人在某个位置

[ 'c' , '_' ,'A' ,'_', '_' , '_']
[ '_' , '_' ,'a' ,'_', '_' , '_']
[ '_' , '_' ,'_' ,'_', 'b' , '_']
[ '_' , '_' ,'_' ,'_', '_' , '_']
[ 'D' , 'd' ,'_' ,'_', '_' , 'B']
[ '_' , '_' ,'_' ,'C', '_' , '_']

输出:像这样的[A:1, B:3, C:8, D:1]


准则:

最靠近自行车的人，将自行车放在第一位。

不能将单个自行车分配给2个人

一辆自行车到一个人的距离永远不会等于同一辆自行车到另一个人的距离。

距离可以相等，但是2个不同的自行车和2个不同的个人

我觉得图形表示可能更有意义



我的方法:

查找Bikes and Person的位置并将其存储在Array中。
person = [[0,2],[4,0],[4,5],[5,3]], bikes = [[0,0],[1,2],[2,4],[4,1]];

因为最短的路径是1，所以开始从Array移除自行车和人
谁将最短路径设为1，并将最短路径递增1。
并将人和自行车存储到结果数组中。

需要继续执行步骤2，直到我们的人员数组为空为止

function findBikesForPeople(grid) {

  let row_length = grid.length;
  let col_length = grid[0].length;
  var bikes = [],
    person = [];

  for (var row = 0; row < row_length; row++) {
    for (var col = 0; col < col_length; col++) {
      if (grid[row][col] === 'B') {
        bikes.push([row, col]);
      }
      if (grid[row][col] === 'P') {
        person.push([row, col]);
      }
    }
  }

  var distances = (bikes, person) => {
    var dist = [];
    person.map((single) => {
      var inner = [];
      bikes.map((bike) => {
        inner.push(check_distance(single, bike));
      })
      dist.push(inner);
    })
    return dist;
  }


  //This isn't right
  var AllocateBikes = (distances) => {
    //var result = [];
    //var min = 1;
    //var increment = 0;
    //  let people = distances.length;
    //let bikeCount = distances[0].length;
    //while (people > 0) {
    //  if (Math.min(...distances[]))
    // }
    return distances;
  }

  function check_distance(a, b) {
    return Math.abs(b[1] - a[1]) + Math.abs(b[0] - a[0]);
  }

  let distance_between = distances(bikes, person);
  console.log(AllocateBikes(distance_between));

}
var grid = [
  ['P', '_', 'B', '_', '_'],
  ['_', '_', '_', '_', 'B'],
  ['_', '_', '_', '_', '_'],
  ['_', 'P', '_', '_', '_'],
  ['_', '_', '_', '_', 'B']
];

findBikesForPeople(grid);

最佳答案

如果我理解正确，那么您就快到了。实际上，您需要做的就是找到人和自行车的所有组合，并测量它们的距离。然后，您可以根据距离对它们进行排序，然后可以遍历它们，并在遇到一个人还没有自行车且自行车仍然免费的组合时，将自行车分配给人们。这将为每个人分配不同的自行车，并首先使用最短的距离。在javascript中，可能类似于:

function findBikesForPeople(grid) {
    var rows = grid.length, cols = grid[0].length;
    var bikes = [], people = [];
    for (var row = 0; row < rows; row++) {
        for (var col = 0; col < cols; col++) {
            if (grid[row][col] === 'B') {
                bikes.push({y: row, x:col});
            }
            if (grid[row][col] === 'P') {
                people.push({y:row, x:col});
            }
        }
    }
    var combis = [];
    for (var p in people) {
        for (var b in bikes) {
            var d = distance(people[p], bikes[b]);
            combis.push({person:p, bike:b, distance:d});
        }
    }
    combis.sort(function(a,b) {return a.distance - b.distance});
    var hasBike = [], isTaken = [], assignment = [];
    for (var c in combis) {
        var person = combis[c].person, bike = combis[c].bike;
        if (!hasBike[person] && !isTaken[bike]) {
            assignment.push({person:person,
                             px:people[person].x, py:people[person].y,
                             bike:bike,
                             bx:bikes[bike].x, by:bikes[bike].y});
            hasBike[person] = true;
            isTaken[bike] = true;
        }
    }
    return assignment;

    function distance(a, b) {
        return Math.abs(b.x - a.x) + Math.abs(b.y - a.y);
    }
}

var grid = [['B', '_', 'P', '_', '_', '_'],
            ['_', '_', 'B', '_', '_', '_'],
            ['_', '_', '_', '_', 'B', '_'],
            ['_', '_', '_', '_', '_', '_'],
            ['P', 'B', '_', '_', '_', 'P'],
            ['_', '_', '_', 'P', '_', '_']];
document.write(JSON.stringify(findBikesForPeople(grid)));

注意:我正在解释代码中显示的网格，其中x =水平，y =垂直，即grid [y] [x]，其中(0,0)是左上 Angular 。