HDU 4756 Install Air Conditioning(树形dp+MST)

题目链接：

HDU 4756 Install Air Conditioning

题意：

给n个点的二维坐标，要将这n个点连通，边权就是两点距离，其中0号节点是根，可能会破坏除根之外的任意两点的边（不能建边），要求将这n个点连通的最坏代价？

数据范围：n≤1000

分析：

上面的黑体字是因为原题中有这句话：

But Tom200 is informed that there are so many wires between two specific dormitories that we cannot set a new high-load wire between these two, otherwise it may have potential risks.

其实这道题的意思和HDU 4126 Genghis Khan the Conqueror是一样的，注意到上面的Trick就好了。枚举破坏的边即可。

时间复杂度：O(n2)。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;
const int MAX_N = 1010;

int T, n, K;
double dis[MAX_N][MAX_N], way[MAX_N], x[MAX_N], y[MAX_N], dp[MAX_N][MAX_N];
int fa[MAX_N], vis[MAX_N];
vector<int> vec[MAX_N];

double GetDis(int i, int j)
{
double xx = x[i] - x[j], yy = y[i] - y[j];
return sqrt(xx * xx + yy * yy);
}

double Prim()
{
double res = 0;
for (int i = 0; i < n; ++i) {
fa[i] = vis[i] = 0;
way[i] = dis[i][0];
}
vis[0] = 1, fa[0] = -1;
for (int i = 1; i < n; ++i) {
double Min = 1e20;
int id;
for (int j = 0; j < n; ++j) {
if (!vis[j] && way[j] < Min) {
Min = way[j];
id = j;
}
}
vis[id] = 1;
res += Min;
if (fa[id] != -1) {
vec[id].push_back(fa[id]);
vec[fa[id]].push_back(id);
}
for (int j = 0; j < n; ++j) {
if (!vis[j] && dis[j][id] < way[j]) {
way[j] = dis[j][id];
fa[j] = id;
}
}
}
return res;
}

double dfs(int root, int u, int p)
{
double Min = 1e20;
for (int i = 0; i < vec[u].size(); ++i) {
int v = vec[u][i];
if (v == p) continue;
double tmp = dfs(root, v, u);
Min = min(Min, tmp);
dp[u][v] = dp[v][u] = min(dp[u][v], tmp);
}
if (p != -1 && p != root) Min = min(Min, dis[root][u]);
return Min;
}

int main()
{
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &K);
for (int i = 0; i < n; ++i) {
vec[i].clear();
scanf("%lf%lf", &x[i], &y[i]);
for (int j = 0; j < i; ++j) {
dis[i][j] = dis[j][i] = GetDis(i, j);
dp[i][j] = dp[j][i] = 1e20;
}
}
double sum = Prim();
for (int i = 0; i < n; ++i) dfs(i, i, -1);
double ans = sum;
for (int i = 1; i < n; ++i) {
for (int j = 1; j < i; ++j) {
if (fa[i] == j || fa[j] == i) {
ans = max(ans, sum - dis[i][j] + dp[i][j]);
}
}
}
printf("%.2lf\n", ans * K);
}
return 0;
}