UVA 1494 - Qin Shi Huang‘s National Road System
题意:秦始皇修路,要求所有道路连通,现在道士徐福可以用法术修一条路,问现在用法术修路的两边的人口数A,除以总修路长度B的最大值A/B是多少
思路:先求出最小生成树,然后利用dfs找出每两点之间的最大权的边的权值,然后在枚举哪两个城市需要法术修路,这样就可以记录下答案最大值
代码:
#include <cstdio> #include <cstring> #include <vector> #include <cmath> #include <algorithm> using namespace std; const int N = 1005; int t, n, parent[N]; struct City { int x, y, p; void read() { scanf("%d%d%d", &x, &y, &p); } } c[N]; struct Edge { int u, v; double val; Edge() {} Edge(int u, int v, double val) { this->u = u; this->v = v; this->val = val; } } e[N * N]; int en; double dis(City a, City b) { int dx = a.x - b.x; int dy = a.y - b.y; return sqrt(dx * dx * 1.0 + dy * dy); } bool cmp(Edge a, Edge b) { return a.val < b.val; } int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]); } vector<Edge> g[N]; double Maxcost[N][N]; void dfs(int s, int u, double Max, int fa) { Maxcost[s][u] = max(Maxcost[s][u], Max); for (int i = 0; i < g[u].size(); i++) { int v = g[u][i].v; if (v == fa) continue; double tmp = max(Max, g[u][i].val); dfs(s, v, tmp, u); } } int main() { scanf("%d", &t); while (t--) { scanf("%d", &n); for (int i = 0; i < n; i++) { c[i].read(); parent[i] = i; g[i].clear(); } en = 0; for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) e[en++] = Edge(i, j, dis(c[i], c[j])); sort(e, e + en, cmp); double sum = 0; for (int i = 0; i < en; i++) { int pu = find(e[i].u); int pv = find(e[i].v); if (pu != pv) { parent[pu] = pv; sum += e[i].val; g[e[i].u].push_back(e[i]); swap(e[i].u, e[i].v); g[e[i].u].push_back(e[i]); } } memset(Maxcost, 0, sizeof(Maxcost)); for (int i = 0; i < n; i++) dfs(i, i, 0, -1); double ans = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { ans = max((c[i].p + c[j].p) * 1.0 / (sum - Maxcost[i][j]), ans); } } printf("%.2lf\n", ans); } return 0; }
UVA 1494 - Qin Shi Huang's National Road System(MST)
时间: 2024-11-06 07:06:25