HYSBZ 1758 重建计划

Description



Input

第一行包含一个正整数N，表示X国的城市个数. 第二行包含两个正整数L和U，表示政策要求的第一期重建方案中修建道路数的上下限 接下来的N-1行描述重建小组的原有方案，每行三个正整数Ai,Bi,Vi分别表示道路(Ai,Bi),其价值为Vi 其中城市由1..N进行标号

Output

输出最大平均估值，保留三位小数

Sample Input

4
2 3
1 2 1
1 3 2
1 4 3

Sample Output

2.500

Hint

20%的数据,N<=5000

30%的数据,N<=100000,原有方案恰好为一条路径

100%的数据,N<=100000,1<=L<=U<=N-1,Vi<=1000000

先二分答案，然后用树分治加单调队列优化搞定。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 2e5 + 10;
int n, l, r, x, y, z;

struct Tree
{
int ft[maxn], nt[maxn], u[maxn], v[maxn], sz;
int mx[maxn], ct[maxn], vis[maxn], SZ;
int FT[maxn], NT[maxn], U[maxn], len[maxn];
int f[maxn], b[maxn], t;
double dp[maxn], g[maxn], a[maxn];
void clear(int n)
{
mx[SZ = sz = t = 0] = INF;
for (int i = 1; i <= n; i++) FT[i] = ft[i] = -1, f[i] = vis[i] = 0;
}
void AddEdge(int x, int y, int z)
{
u[sz] = y; v[sz] = z; nt[sz] = ft[x]; ft[x] = sz++;
u[sz] = x; v[sz] = z; nt[sz] = ft[y]; ft[y] = sz++;
}
int dfs(int x, int fa, int sum)
{
int y = mx[x] = (ct[x] = 1) ^ 1;
for (int i = ft[x]; i != -1; i = nt[i])
{
if (vis[u[i]] || u[i] == fa) continue;
int z = dfs(u[i], x, sum);
ct[x] += ct[u[i]];
mx[x] = max(mx[x], ct[u[i]]);
y = mx[y] < mx[z] ? y : z;
}
mx[x] = max(mx[x], sum - ct[x]);
return mx[x] < mx[y] ? x : y;
}
int getdep(int x, int fa, int dep)
{
int ans = dep;
for (int i = ft[x]; i != -1; i = nt[i])
{
if (vis[u[i]] || u[i] == fa) continue;
ans = max(ans, getdep(u[i], x, dep + 1));
}
return ans;
}
int build(int x, int sum)
{
int y = dfs(x, -1, sum);
vis[y] = 1; len[y] = getdep(y, -1, 0);
for (int i = ft[y]; i != -1; i = nt[i])
{
if (vis[u[i]]) continue;
int z = build(u[i], ct[u[i]] > ct[y] ? sum - ct[y] : ct[u[i]]);
U[SZ] = z; NT[SZ] = FT[y]; FT[y] = SZ++;
}
vis[y] = 0;
return y;
}
void get(int x, int fa, int dep, double len, double c)
{
if (f[dep] != t) g[dep] = len, f[dep] = t; else g[dep] = max(g[dep], len);
for (int i = ft[x]; i != -1; i = nt[i])
{
if (vis[u[i]] || u[i] == fa) continue;
get(u[i], x, dep + 1, len + v[i] - c, c);
}
}
bool work(int x, double c)
{
bool flag = false;
dp[0] = 0;
for (int i = 1; i <= len[x]; i++) dp[i] = -INF;
for (int i = ft[x]; i != -1; i = nt[i])
{
if (vis[u[i]]) continue;
++t; get(u[i], x, 1, v[i] - c, c);
int h = 0, q = 1, k;
for (k = min(r - 1, len[x]); k >= l - 1 && k >= 0; k--)
{
if (dp[k] == -INF) continue;
while (q <= h&&dp[k] > a[h]) --h;
a[++h] = dp[k], b[h] = k;
}
k = l - 1;
for (int j = 1; f[j] == t; j++)
{
while (q <= h && (b[q] + j < l || b[q] + j>r)) ++q;
if (q <= h&&a[q] + g[j] >= 0) { flag = true; break; }
if (--k >= 0)
{
if (k > len[x] || dp[k] == -INF) continue;
while (q <= h&&dp[k] > a[h]) --h;
a[++h] = dp[k], b[h] = k;
}
}
for (int j = 1; f[j] == t; j++) dp[j] = max(dp[j], g[j]);
}

vis[x] = 1;
for (int i = FT[x]; i != -1 && !flag; i = NT[i]) flag |= work(U[i], c);
vis[x] = 0;
return flag;
}
}solve;

int main()
{
while (scanf("%d%d%d", &n, &l, &r) != EOF)
{
solve.clear(n);
double h = -INF, q = INF;
for (int i = 1; i < n; i++)
{
scanf("%d%d%d", &x, &y, &z);
solve.AddEdge(x, y, z);
q = min(q, 1.0*z); h = max(h, 1.0*z);
}
int root = solve.build(1, n);
while (h - q > 1e-4)
{
double m = (q + h) / 2;
if (solve.work(root, m)) q = m; else h = m;
}
printf("%.3lf\n", q);
}
return 0;
}