LeetCode:1584. Min Cost to Connect All Points連線所有點的最小費用(C語言)
阿新 • • 發佈:2021-02-04
技術標籤:LeetCode
題目描述:
給你一個points 陣列,表示 2D 平面上的一些點,其中 points[i] = [xi, yi] 。
連線點 [xi, yi] 和點 [xj, yj] 的費用為它們之間的 曼哈頓距離 :|xi - xj| + |yi - yj| ,其中 |val| 表示 val 的絕對值。
請你返回將所有點連線的最小總費用。只有任意兩點之間 有且僅有 一條簡單路徑時,才認為所有點都已連線。
示例 1:
輸入:points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
輸出:20
解釋:
我們可以按照上圖所示連線所有點得到最小總費用,總費用為 20 。
示例 2:
輸入:points = [[3,12],[-2,5],[-4,1]]
輸出:18
示例 3:
輸入:points = [[0,0],[1,1],[1,0],[-1,1]]
輸出:4
示例 4:
輸入:points = [[-1000000,-1000000],[1000000,1000000]]
輸出:4000000
示例 5:
輸入:points = [[0,0]]
輸出:0
提示:
1 <= points.length <= 1000
-106 <= xi, yi <= 106
所有點 (xi, yi) 兩兩不同。
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/min-cost-to-connect-all-points
解答:
void swap(int* a, int* b) {
int tmp = *a;
*a = *b, *b = tmp;
}
struct Edge {
int len, x, y;
};
int cmp(struct Edge* a, struct Edge* b) {
return a->len - b->len;
}
int find(int* f, int x) {
return f[x] == x ? x : (f[x] = find(f, f[x] ));
}
int unionSet(int* f, int* rank, int x, int y) {
int fx = find(f, x), fy = find(f, y);
if (fx == fy) {
return false;
}
if (rank[fx] < rank[fy]) {
swap(&fx, &fy);
}
rank[fx] += rank[fy];
f[fy] = fx;
return true;
}
int minCostConnectPoints(int** points, int pointsSize, int* pointsColSize) {
int n = pointsSize;
struct Edge edges[(n + 1) * n / 2];
int edgesSize = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
edges[edgesSize].x = i;
edges[edgesSize].y = j;
edges[edgesSize++].len = fabs(points[i][0] - points[j][0]) + fabs(points[i][1] - points[j][1]);
}
}
qsort(edges, edgesSize, sizeof(struct Edge), cmp);
int f[n], rank[n];
for (int i = 0; i < n; i++) {
f[i] = i;
rank[i] = 1;
}
int ret = 0, num = 1;
for (int i = 0; i < edgesSize; i++) {
if (unionSet(f, rank, edges[i].x, edges[i].y)) {
ret += edges[i].len;
num++;
if (num == n) {
break;
}
}
}
return ret;
}
執行結果:
Notes:
參考官方文件:https://leetcode-cn.com/problems/min-cost-to-connect-all-points/solution/lian-jie-suo-you-dian-de-zui-xiao-fei-yo-kcx7/