这道题是最小生成树板子题
可以用并查集实现,贪心排序边权
讲一个二元组放在一个vector容器里面,其中的元素为<weight,<u,v>>对应<int,<int,int> >类型,第一个参数代表边权的大小,后面的为两个点u,v,然后按照第一个值边权从小到大排序,然后用并查集实现是否连通,从而实现最小生成树
代码有点套娃(
class Solution {
private:
int fa[1007];
int cnt = 0;
typedef pair<int, pair<int, int> > pir;
void init() {
for (int i = 0; i < 1007; i++) fa[i] = i;
}
int _find(int u) {
if (fa[u] == u) return u;
else return fa[u] = _find(fa[u]);
}
public:
int minCostConnectPoints(vector<vector<int>> &points) {
init();
vector<pir> vt;
int n = points.size();
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int wei = abs(points[i][0] - points[j][0]);
wei += abs(points[i][1] - points[j][1]);
vt.push_back({wei, {i, j}});
}
}
sort(vt.begin(), vt.end(), [](pir p1, pir p2) {
return p1.first < p2.first;
});
long long out = 0;
int lim = vt.size();
for (int i = 0; i < lim; i++) {
int u = vt[i].second.first;
int v = vt[i].second.second;
int fau = _find(fa[u]);
int fav = _find(fa[v]);
if (fau == fav) continue;
else {
fa[fau] = fav;
out += vt[i].first;
}
}
return out;
}
};