题意：给定上一棵树，然后每条边有一个权值，然后每个点到 1 的距离有两种，第一种是直接回到1，花费是 dist(1, i)^2，还有另一种是先到另一个点 j，然后两从 j 向1走，当然 j 也可以再向 k，一直到1，但经过一个点，那么就会出多一个花费 p，问你每个点到 1 的最小距离的最大值是多少。
析：很容易想到状态方程是 dp[i] = min{ dp[j] + (dist(1, i) - dist(1, j))^2 + P } dist(1, i) 表示 1 到 i 的距离。但可以是斜率进行优化， 这样的话，时间复杂度就小了，只不是树上的而已。
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 <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 50;
const int maxm = 1e6 + 5;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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 bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int to, dist, next;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
void addEdge(int u, int v, int dist){
edge[cnt].to = v;
edge[cnt].dist = dist;
edge[cnt].next = head[u];
head[u] = cnt++;
}
LL dp[maxn];
struct Node{
int num, pos, time;
};
stack<Node> st;
int q[maxn], dfs_cnt, fro, rear;
LL sum[maxn];
LL DP(int i, int j){
return dp[j] + sqr(sum[i]-sum[j]) + m;
}
LL UP(int j, int k){
return dp[j] + sqr(sum[j]) - dp[k] - sqr(sum[k]);
}
LL DOWN(int j, int k){
return 2LL * (sum[j] - sum[k]);
}
void dfs(int u, int fa){
int ti = ++dfs_cnt;
while(fro + 1 < rear && UP(q[fro+2], q[fro+1]) <= sum[u] * DOWN(q[fro+2], q[fro+1])) ++fro;
dp[u] = DP(u, q[fro+1]);
while(fro + 1 < rear && UP(u, q[rear]) * DOWN(u, q[rear-1]) <= UP(u, q[rear-1]) * DOWN(u, q[rear])){
Node tmp;
tmp.pos = rear;
tmp.num = q[rear];
tmp.time = dfs_cnt;
st.push(tmp);
--rear;
}
q[++rear] = u;
int nowfro = fro, nowrear = rear;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
sum[v] = sum[u] + edge[i].dist;
fro = nowfro; rear = nowrear;
while(!st.empty()){
Node tmp = st.top();
if(tmp.time <= ti) break;
q[tmp.pos] = tmp.num;
st.pop();
}
dfs(v, u);
}
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
ms(head, -1); cnt = 0;
for(int i = 1; i < n; ++i){
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
addEdge(u, v, c);
addEdge(v, u, c);
}
dp[1] = -m;
q[++rear] = 1;
dfs_cnt = 0;
for(int i = head[1]; ~i; i = edge[i].next){
int v = edge[i].to;
fro = 0, rear = 1;
while(!st.empty()) st.pop();
sum[v] = sum[1] + edge[i].dist;
dfs(v, 1);
}
LL ans = 0;
for(int i = 1; i <= n; ++i) ans = max(ans, dp[i]);
printf("%I64d
", ans);
}
return 0;
}