题目链接

Problem Description

Once upon a time, in the mystical continent, there is a frog kingdom, ruled by the oldest frog, the Elder. The kingdom consists of N cities, numbered from east to west. The 1-th city, which is located to the east of others, is the capital. Each city, except the capital, links none or several cities to the west, and exactly one city to the east.
There are some significant news happening in some cities every day. The Elder wants to know them as soon as possible. So, that is the job of journalist frogs, who run faster than any other frog. Once some tremendous news happen in a city, the journalist in that city would take the message and run to the capital. Once it reach another city, it can either continue running, or stop at that city and let another journalist to transport. The journalist frogs are too young and simple to run a long distance efficiently. As a result, it takes L2 time for them to run through a path of length L. In addition, passing message requires P time for checking the message carefully, because any mistake in the message would make the Elder become extremely angry.
Now you are excited to receive the task to calculate the maximum time of sending a message from one of these cities to the capital.

Input

The first line of input contains an integer t, the number of test cases. t test cases follow. For each test case, in the first line there are two integers N (N ≤ 100000) and P (P ≤ 1000000). In the next N-1 lines, the i-th line describes the i-th road, a line with three integers u,v,w denotes an edge between the u-th city and v-th city with length w(w ≤ 100).

Output

For each case, output the maximum time.

Sample Input

3 6 10 1 2 4 2 3 5 1 4 3 4 5 3 5 6 3 6 30 1 2 4 2 3 5 1 4 3 4 5 3 5 6 3 6 50 1 2 4 2 3 5 1 4 3 4 5 3 5 6 3

Sample Output

51 75 81

Hint

In the second case, the best transportation time is: • The 2-th city: 16 = 4^2 • The 3-th city: 72 = 4^2 + 30 + 5^2 • The 4-th city: 9 = 3^2 • The 5-th city: 36 = (3 + 3)^2 • The 6-th city: 75 = (3 + 3)^2 +30 + 3^2 Consequently, the news in the 6-th city requires most time to reach the capital.

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e5+10;
const int mod=100000000;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;
struct node
{int v,w,next;
} maps[maxn<<1];
int head[maxn],tot;
int que[maxn],p,n,vis[maxn];
ll dis[maxn],dp[maxn],ans;
void init()
{me(head,-1),me(vis,0),me(dis,0);tot=0,ans=0,que[0]=0;
}
void add(int u,int v,int w)
{maps[tot].v=v;maps[tot].w=w;maps[tot].next=head[u];head[u]=tot++;
}
ll DY(int j,int k)
{return dp[j]+dis[j]*dis[j]-(dp[k]+dis[k]*dis[k]);
}
ll DX(int j,int k)
{return 2*(dis[j]-dis[k]);
}
ll DP(int i,int j)
{return dp[j]+(dis[i]-dis[j])*(dis[i]-dis[j])+p;//状态转移方程 dp[i]=min(dp[i],dp[j]+(dis[i]-dis[j])*(dis[i]-dis[j])+p);
}
void dfs(int u)//计算所有点到根节点的距离
{vis[u]=1;for(int i=head[u]; i!=-1; i=maps[i].next){int v=maps[i].v;if(!vis[v]){dis[v]=dis[u]+maps[i].w;dfs(v);}}
}
void DFS(int u,int fa,int s,int t)
{int pre=-1;while(s<t&&DY(que[s+1],que[s])<=dis[u]*DX(que[s+1],que[s]))s++;dp[u]=min(dp[u],DP(u,que[s]));while(s<t&&DY(que[t],que[t-1])*DX(u,que[t])>=DY(u,que[t])*DX(que[t],que[t-1]))t--;pre=que[++t];que[t]=u;ans=max(ans,dp[u]);for(int i=head[u]; i!=-1; i=maps[i].next){int v=maps[i].v;if(v!=fa)DFS(v,u,s,t);}if(pre!=-1)//因为u的子节点不止一个，下面还有其他节点。que[t]=pre;
}
int main()
{int t;scanf("%d",&t);while(t--){init();scanf("%d%d",&n,&p);for(int i=1; i<n; i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);add(u,v,w),add(v,u,w);}dfs(1);for(int i=1; i<=n; i++)dp[i]=dis[i]*dis[i];DFS(1,0,1,1);printf("%lld\n",ans);}return 0;
}

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HDU 5956 The Elder(树型DP+斜率优化)

题目链接

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