Qin Shi Huang's National Road System HDU - 4081（树形dp+最小生成树）

Qin Shi Huang's National Road System

HDU - 4081

感觉这道题和hdu4756很像...

求最小生成树里面删去一边E1 再加一边E2 求该边两顶点权值和除以（最小生成树-E1）的最大值

其中（最小生成树-E1）必须是最小的

先跑一遍prim 跑完之后在最小生成树里面dp

dp[i][j] = i到j的路径中最大的那条边 最小生成树减去dp[i][j]肯定会最小

代码如下

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

const double inf = 0x3f3f3f3f3f;
const int maxn = ;
struct Point {
    double x, y;
    int n;
} point[maxn];
struct Edge {
    int to;
    int next;
} edge[maxn<<];
int n, cnt, head[maxn], pre[maxn];
double dis[maxn][maxn], lowc[maxn], sum, dp[maxn][maxn];
bool vis[maxn];
inline void addedge(int u, int v) {
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
inline double Distance(const Point& lhs, const Point& rhs) {
    return sqrt((lhs.x - rhs.x) * (lhs.x - rhs.x) + (lhs.y - rhs.y) * (lhs.y - rhs.y));
}
void prim() {
    sum = 0.0;
    memset(vis, , sizeof(vis));
    memset(pre, , sizeof(pre));
    for (int i = ; i < n; i++) lowc[i] = dis[][i];
    vis[] = true;
    for (int i = ; i < n; i++) {
        double minc = inf;
        int p = -;
        for (int j = ; j < n; j++) {
            if (!vis[j] && minc > lowc[j]) {
                minc = lowc[j];
                p = j;
            }
        }
        sum += minc;
        vis[p] = true;
        addedge(p, pre[p]);
        addedge(pre[p], p);
        for (int j = ; j < n; j++) {
            if (!vis[j] && lowc[j] > dis[p][j]) {
                lowc[j] = dis[p][j];
                pre[j] = p;
            }
        }
    }
}
void dfs(int u, int root) {
    vis[u] = true;
    for (int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].to;
        if (!vis[v]) {
            dp[root][v] = max(dp[root][u], dis[u][v]);
            dfs(v, root);
        }
    }
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (int i = ; i < n; i++) {
            scanf("%lf%lf%d", &point[i].x, &point[i].y, &point[i].n);
        }
        for (int i = ; i < n; i++) {
            for (int j = i + ; j < n; j++) {
                dis[i][j] = dis[j][i] = Distance(point[i], point[j]);
            }
        }
        memset(head, -, sizeof(head));
        cnt = ;
        prim();
        for (int i = ; i < n; i++) {
            memset(vis, ,sizeof(vis));
            dfs(i, i);
        }
        double ans = ;
        for (int i = ; i < n; i++) {
            for (int j = i + ; j < n; j++) {
                double temp = sum - dp[i][j];
                temp = (point[i].n + point[j].n) / temp;
                ans = max(ans, temp);
            }
        }
        printf("%.2f\n", ans);
    }
    return ;
}