HDU 1875 畅通工程再续——最小生成树

HDU 1875 畅通工程再续——最小生成树

先遍历所有节点求出所有节点间的距离，然后用Kruskal算法求最小生成树。

#include #include #include #include #include using namespace std;const int maxn = 110;int par[maxn], ran[maxn];void init(int n) { for (int i = 0; i <= n; i++) { par[i] = i, ran[i] = 0; }}int seek(int x) { if (par[x] == x) return x; else return par[x] = seek(par[x]);}void unite(int x, int y) { x = seek(x), y = seek(y); if (x == y) return; if (ran[x] == ran[y]) par[x] = y; else { par[y] = x; if (ran[x] == ran[y]) ran[x]++; }}struct Edge { int u, v; double cost; bool operator < (const Edge &another) const { return cost < another.cost; }}edge[maxn*maxn];struct Node { int x, y;}node[maxn];double cal(int x1, int y1, int x2, int y2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}int main(){ int T, n; scanf("%d", &T); while (T--) { scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d %d", &node[i].x, &node[i].y); } int edge_cnt = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { double temp = cal(node[i].x, node[i].y, node[j].x, node[j].y); if (temp < 10 || temp > 1000) continue; edge_cnt++; edge[edge_cnt].u = i, edge[edge_cnt].v = j, edge[edge_cnt].cost = temp * 100.0; } } sort(edge + 1, edge + 1 + edge_cnt); init(n); int cnt = 0; double ans = 0.0; for (int i = 1; i <= edge_cnt; i++) { if (cnt == n - 1) break; if (seek(edge[i].u) != seek(edge[i].v)) { cnt++; ans += edge[i].cost; unite(edge[i].u, edge[i].v); } } if (cnt == n - 1) printf("%.1lf\n", ans); else printf("oh!\n"); } return 0;}

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