leetcode1066 校园自行车分配II

题目描述

在由 2D 网格表示的校园里有 n 位工人（worker）和 m 辆自行车（bike），n <= m。所有工人和自行车的位置都用网格上的 2D 坐标表示。

我们为每一位工人分配一辆专属自行车，使每个工人与其分配到的自行车之间的曼哈顿距离最小化。

p1 和 p2 之间的曼哈顿距离为 Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|。

返回每个工人与分配到的自行车之间的曼哈顿距离的最小可能总和。

示例1:

输入：workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
输出：6
解释：
自行车 0 分配给工人 0，自行车 1 分配给工人 1 。分配得到的曼哈顿距离都是 3, 所以输出为 6 。

示例2:

输入：workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
输出：4
解释：
先将自行车 0 分配给工人 0，再将自行车 1 分配给工人 1（或工人 2），自行车 2 给工人 2（或工人 1）。如此分配使得曼哈顿距离的总和为 4。

提示:

1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
2. 所有工人和自行车的位置都不相同。
3. 1 <= workers.length <= bikes.length <= 10

解题思路

暴力搜索

很自然我们想到可以使用两个bitmap: workerBitmap 和 bikeBitmap 来标识哪些工人和自行车已经匹配，下一步是从两个bitmap中选取没有被占用的位置进行匹配，暴力搜索如下，在运行到11的用例的时候判断超时失败。

int getBikeDistance(int *worker, int *bike)
{return abs(worker[0] - bike[0]) + abs(worker[1] - bike[1]);
}

int helper(int** workers, int* workersBitmap, int workersSize, int** bikes, int* bikesBitmap, int bikesSize)
{int assignWorkerSize = 0;for (int i = 0; i < workersSize; i++) {if (workersBitmap[i]) {assignWorkerSize ++;}}
if (assignWorkerSize == workersSize) {return 0;}
int minDist = INT_MAX;for (int i = 0; i < workersSize; i++) {if (workersBitmap[i] == 1) {continue;}
workersBitmap[i] = 1;for (int j = 0; j < bikesSize; j++) {if (bikesBitmap[j] == 1) {continue;}
bikesBitmap[j] = 1;
int dist = getBikeDistance(workers[i], bikes[j]);dist += helper(workers, workersBitmap, workersSize, bikes, bikesBitmap, bikesSize);
minDist = dist > minDist ? minDist : dist;bikesBitmap[j] = 0;}workersBitmap[i] = 0;}
return minDist;
}

int assignBikes(int** workers, int workersSize, int* workersColSize, int** bikes, int bikesSize, int* bikesColSize)
{int workersBitmap[workersSize];int bikesBitmap[bikesSize];
memset(workersBitmap, 0, sizeof(workersBitmap));memset(bikesBitmap, 0, sizeof(bikesBitmap));
return helper(workers, workersBitmap, workersSize, bikes, bikesBitmap, bikesSize);
}

BackTracing+位压缩

在暴力解法中，我们每一次从worker集合中随机取一个worker和bike集合中的一个进行匹配，最后worker的bitmap都是全1的，导致我们随机选取worker的过程中会有大量的重复操作(匹配结果都相同)。

我们考虑按序使用每一个worker去和bike集中的一个进行匹配，继续保留bikeBitmap。所以如果第 i 个worker匹配到了第 j 个bike之后，实际上前i个workerBitmap中的元素已经全部置1，匹配第i + 1的时候，我们可以从bikeBitmap中挑选0的bike进行匹配。最后的最优解就是第i个元素匹配上的bike的曼哈顿距离加上任意选择一个后从第 i + 1个worker开始和剩余bike的和的最小值。

观察到题目限制，工人和单车最多只有10个，所以我们可以将bike的状态压缩到一个INT32的数字中，然后从前往后回溯。提交显示AC 992ms。

int getBikeDistance(int *worker, int *bike)
{return abs(worker[0] - bike[0]) + abs(worker[1] - bike[1]);
}int helper(int** workers, int workerIndex, int workersSize, int** bikes, int bikeState, int bikesSize)
{if (workerIndex == workersSize) {return 0;}int minDist = INT_MAX;for (int bikeIndex = 0; bikeIndex < bikesSize; bikeIndex++) {if ((bikeState & (1 << bikeIndex)) != 0) {continue;}int dist = getBikeDistance(workers[workerIndex], bikes[bikeIndex]) + helper(workers, workerIndex + 1, workersSize, bikes, bikeState | (1 << bikeIndex), bikesSize);minDist = dist > minDist ? minDist : dist;}return minDist;
}int assignBikes(int** workers, int workersSize, int* workersColSize, int** bikes, int bikesSize, int* bikesColSize)
{return helper(workers, 0, workersSize, bikes, 0, bikesSize);
}

BackTracking+位压缩+记忆化搜索去重

在上面的方法中，我们已经通过回溯方法通过了此问题。我们画出问题决策树，可以看到选择不同的bike之间存在大量的重复。

我们可以将节点看作是问题的最优解，使用一个Memo数组记忆化去重。AC通过时间4ms。

#define MAX_BIKE_STATE 1024
int g_workerBikeDistance[MAX_BIKE_STATE];

int workerBikeDistance(int *worker, int *bike)
{return abs(worker[0] - bike[0]) + abs(worker[1] - bike[1]);
}

int helper(int** workers, int workerIndex, int workersSize, int** bikes, int bikeState, int bikesSize)
{if (workerIndex == workersSize) {return 0;}
int minDist = INT_MAX;for (int bikeIndex = 0; bikeIndex < bikesSize; bikeIndex++) {if ((bikeState & (1 << bikeIndex)) != 0) {continue;}
int dist = workerBikeDistance(workers[workerIndex], bikes[bikeIndex]);if (g_workerBikeDistance[bikeState | (1 << bikeIndex)] == -1) {dist += helper(workers, workerIndex + 1, workersSize, bikes, bikeState | (1 << bikeIndex), bikesSize);} else {dist += g_workerBikeDistance[bikeState | (1 << bikeIndex)];}minDist = minDist > dist ? dist : minDist;}g_workerBikeDistance[bikeState] = minDist;return minDist;
}

int assignBikes(int** workers, int workersSize, int* workersColSize, int** bikes, int bikesSize, int* bikesColSize)
{memset(g_workerBikeDistance, -1, sizeof(g_workerBikeDistance));
return helper(workers, 0, workersSize, bikes, 0, bikesSize);
}