HDU 4756 Install Air Conditioning (MST+树形DP)
2024-10-20 16:01:59
题意:n-1个宿舍,1个供电站,n个位置每两个位置都有边相连,其中有一条边不能连,求n个位置连通的最小花费的最大值。
析:因为要连通,还要权值最小,所以就是MST了,然后就是改变一条边,然后去找出改变哪条能使得总花费最大,dp[i][j] 表示那条边左边的 i 和右边的 j,
最短距离,然后枚举MST里面的每条边,就能知道哪是最大了,注意 供电站和宿舍之间的边不能考虑的。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Point{
double x, y;
};
struct Edge{
int to, next;
};
Edge edge[maxn<<1];
Point a[maxn];
double dist[maxn][maxn], lowc[maxn], dp[maxn][maxn];
bool vis[maxn], is_tree[maxn][maxn];
int pre[maxn], head[maxn];
int cnt;
double sum, ans; double Distan(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 add(int u, int v){
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
} void Prim(){
sum = 0.0;
memset(vis, false, sizeof vis);
memset(pre, 0, sizeof pre);
for(int i = 1; i < n; ++i) lowc[i] = dist[0][i];
vis[0] = true; for(int i = 1; i < n; ++i){
double minc = inf;
int p = -1;
for(int j = 0; j < n; ++j)
if(!vis[j] && minc > lowc[j]) minc = lowc[j], p = j;
sum += minc;
vis[p] = true;
add(p, pre[p]);
add(pre[p], p);
for(int j = 0; j < n; ++j)
if(!vis[j] && lowc[j] > dist[p][j])
lowc[j] = dist[p][j], pre[j] = p;
}
} double dfs(int u, int fa, int root){
double ans = fa == root ? inf : dist[root][u];
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
double tmp = dfs(v, u, root);
ans = min(ans, tmp);
dp[u][v] = dp[v][u] = min(dp[u][v], tmp);
}
return ans;
} void dfs1(int u, int fa){
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
if(fa) ans = max(ans, sum-dist[u][v]+dp[u][v]);
dfs1(v, u);
}
} int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i) scanf("%lf %lf", &a[i].x, &a[i].y);
for(int i = 0; i < n; ++i)
for(int j = i+1; j < n; ++j)
dist[i][j] = dist[j][i] = Distan(a[i], a[j]); cnt = 0;
memset(head, -1, sizeof head);
Prim(); for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j) dp[i][j] = inf;
for(int i = 0; i < n; ++i) dfs(i, -1, i); ans = sum;
dfs1(0, 0);
ans *= m * 1.0;
printf("%.2f\n", ans);
}
return 0;
}
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