洛谷—— P2934 [USACO09JAN]安全出行Safe Travel || COGS ——279|| BZOJ—

https://www.luogu.org/problem/show?pid=2934

题目描述

Gremlins have infested the farm. These nasty, ugly fairy-like

creatures thwart the cows as each one walks from the barn (conveniently located at pasture_1) to the other fields, with cow_i traveling to from pasture_1 to pasture_i. Each gremlin is personalized and knows the quickest path that cow_i normally takes to pasture_i. Gremlin_i waits for cow_i in the middle of the final cowpath of the quickest route to pasture_i, hoping to harass cow_i.

Each of the cows, of course, wishes not to be harassed and thus chooses an at least slightly different route from pasture_1 (the barn) to pasture_i.

Compute the best time to traverse each of these new not-quite-quickest routes that enable each cow_i that avoid gremlin_i who is located on the final cowpath of the quickest route from pasture_1 to

pasture_i.

As usual, the M (2 <= M <= 200,000) cowpaths conveniently numbered 1..M are bidirectional and enable travel to all N (3 <= N <= 100,000) pastures conveniently numbered 1..N. Cowpath i connects pastures a_i (1 <= a_i <= N) and b_i (1 <= b_i <= N) and requires t_i (1 <= t_i <= 1,000) time to traverse. No two cowpaths connect the same two pastures, and no path connects a pasture to itself (a_i != b_i). Best of all, the shortest path regularly taken by cow_i from pasture_1 to pasture_i is unique in all the test data supplied to your program.

By way of example, consider these pastures, cowpaths, and [times]:

1--[2]--2-------+

| | | [2] [1] [3]

| | | +-------3--[4]--4

TRAVEL BEST ROUTE BEST TIME LAST PATH

p_1 to p_2 1->2 2 1->2

p_1 to p_3 1->3 2 1->3

p_1 to p_4 1->2->4 5 2->4

When gremlins are present:

TRAVEL BEST ROUTE BEST TIME AVOID

p_1 to p_2 1->3->2 3 1->2

p_1 to p_3 1->2->3 3 1->3

p_1 to p_4 1->3->4 6 2->4

For 20% of the test data, N <= 200.

For 50% of the test data, N <= 3000.

TIME LIMIT: 3 Seconds

MEMORY LIMIT: 64 MB

Gremlins最近在农场上泛滥,它们经常会阻止牛们从农庄(牛棚_1)走到别的牛棚(牛_i的目的地是牛棚_i).每一个gremlin只认识牛_i并且知道牛_i一般走到牛棚_i的最短路经.所以它们在牛_i到牛棚_i之前的最后一条牛路上等牛_i. 当然,牛不愿意遇到Gremlins,所以准备找一条稍微不同的路经从牛棚_1走到牛棚_i.所以,请你为每一头牛_i找出避免gremlin_i的最短路经的长度. 和以往一样, 农场上的M (2 <= M <= 200,000)条双向牛路编号为1..M并且能让所有牛到达它们的目的地, N(3 <= N <= 100,000)个编号为1..N的牛棚.牛路i连接牛棚a_i (1 <= a_i <= N)和b_i (1 <= b_i <= N)并且需要时间t_i (1 <=t_i <= 1,000)通过. 没有两条牛路连接同样的牛棚,所有牛路满足a_i!=b_i.在所有数据中,牛_i使用的牛棚_1到牛棚_i的最短路经是唯一的.

输入输出格式

输入格式：

Line 1: Two space-separated integers: N and M
Lines 2..M+1: Three space-separated integers: a_i, b_i, and t_i

输出格式：

Lines 1..N-1: Line i contains the smallest time required to travel from pasture_1 to pasture_i+1 while avoiding the final cowpath of the shortest path from pasture_1 to pasture_i+1. If no such path exists from pasture_1 to pasture_i+1, output -1 alone on the line.

输入输出样例

输入样例#1：

输出样例#1：

说明

感谢 karlven 提供翻译。

正解最短路径生成树+树剖（并查集可水过~、、、）

先用Dijkstra计算dis 以及最短路径的最后一条边，

对于这两点u,v,(无向边看做两条有向边)求出lca（u，v），，

则对于lca--v这条路径每个点,都可以由u到达，路径上的上的dis[x]'=dis[u]+dis[v]+edge[i].dis-dis[x](画图意会)

用树剖（并查集）维护最小值

 #include <algorithm>

 #include <cstdio>

 #include <queue>

 using namespace std;

 const int INF(0x3f3f3f3f);

 const int M(6e5+);

 const int N(4e5+);

 int n,m;

 int head[N],sumedge=;

 struct Edge

 {

     int u,v,next,w;

     Edge(int u=,int v=,int next=,int w=):

         u(u),v(v),next(next),w(w){}

 }edge[M<<];

 inline void ins(int u,int v,int w)

 {

     edge[++sumedge]=Edge(u,v,head[u],w);

     head[u]=sumedge;

 }

 struct Node

 {

     int id,dis;

     bool operator < (const Node &x) const

     {

         return dis>x.dis;

     }

 };

 priority_queue<Node>que;

 bool vis[N],mark[N];

 int dis[N],pre[N];

 inline void Dijkstra()

 {

     for(int i=;i<=n;i++) dis[i]=INF;

     dis[]=;

     que.push((Node){,});

     for(;!que.empty();)

     {

         int u=que.top().id;

         que.pop();

         if(vis[u]) continue;

         vis[u]=;

         for(int i=head[u];i;i=edge[i].next)

         {

             int v=edge[i].v;

             if(dis[v]>dis[u]+edge[i].w)

             {

                 dis[v]=dis[u]+edge[i].w;

                 mark[pre[v]]=;

                 mark[i]=;

                 pre[v]=i;

                 que.push((Node){v,dis[v]});

             }

         }

     }

 }

 int size[N],deep[N],dad[N],top[N],son[N],cnt,id[N],dfn[N];

 void DFS(int x,int fa,int deepth)

 {

     size[x]=; deep[x]=deepth; dad[x]=fa;

     for(int i=head[x];i;i=edge[i].next)

         if(mark[i])

         {

             int v=edge[i].v;

             if(fa==v) continue;

             DFS(v,x,deepth+);

             size[x]+=size[v];

             if(size[son[x]]<size[v]) son[x]=v;

         }

 }

 void DFS_(int x,int Top)

 {

     id[x]=++cnt; dfn[cnt]=x;

     top[x]=Top;

     if(son[x]) DFS_(son[x],Top);

     for(int i=head[x];i;i=edge[i].next)

         if(mark[i])

         {

             int v=edge[i].v;

             if(dad[x]!=v&&son[x]!=v) DFS_(v,v);

         }

 }

 int LCA(int x,int y)

 {

     for(;top[x]!=top[y];x=dad[top[x]])

         if(deep[top[x]]<deep[top[y]]) swap(x,y);

     return deep[x]<deep[y]?x:y;

 }

 struct Tree

 {

     int l,r,minn,flag;

 }tr[N<<];

 #define lc (now<<1)

 #define rc (now<<1|1)

 #define mid (tr[now].l+tr[now].r>>1)

 inline void Tree_up(int now)

 {

     tr[now].minn=min(tr[lc].minn,tr[rc].minn);

 }

 void Tree_build(int now,int l,int r)

 {

     tr[now].l=l; tr[now].r=r;

     tr[now].minn=INF;

     tr[now].flag=INF;

     if(l==r) return ;

     Tree_build(lc,l,mid);

     Tree_build(rc,mid+,r);

 }

 inline void Tree_down(int now)

 {

     if(tr[now].flag==INF) return ;

     tr[lc].flag=min(tr[lc].flag,tr[now].flag);

     tr[rc].flag=min(tr[rc].flag,tr[now].flag);

     tr[lc].minn=min(tr[lc].minn,tr[lc].flag);

     tr[rc].minn=min(tr[rc].minn,tr[rc].flag);

 }

 void Tree_change(int now,int l,int r,int x)

 {

     if(tr[now].l==l&&tr[now].r==r)

     {

         tr[now].minn=min(tr[now].minn,x);

         tr[now].flag=min(tr[now].flag,x);

         return ;

     }

     Tree_down(now);

     if(r<=mid) Tree_change(lc,l,r,x);

     else if(l>mid) Tree_change(rc,l,r,x);

     else Tree_change(lc,l,mid,x),Tree_change(rc,mid+,r,x);

     Tree_up(now);

 }

 int Tree_query(int now,int to)

 {

     if(tr[now].l==tr[now].r) return tr[now].minn;

     Tree_down(now);

     if(to<=mid) return Tree_query(lc,to);

     else return Tree_query(rc,to);

 }

 inline void List_change(int u,int v,int w)

 {

     int fx=top[u] ;

     while(deep[fx]>deep[v])

     {

         Tree_change(,id[top[u]],id[u],w);

         u=dad[fx],fx=top[u];

     }

     if(u!=v) Tree_change(,id[v]+,id[u],w);

 }

 int main()

 {

     freopen("travel.in","r",stdin);

     freopen("travel.out","w",stdout);

     scanf("%d%d",&n,&m);

     for(int a,b,t,i=;i<=m;i++)

     {

         scanf("%d%d%d",&a,&b,&t);

         ins(a,b,t); ins(b,a,t);

     }

     Dijkstra();

     DFS(,,);

     DFS_(,);

     Tree_build(,,cnt);

     for(int lca,u,v,i=;i<=sumedge;i+=)

     {

         u=edge[i].u,v=edge[i].v,lca=LCA(u,v);

         if(!mark[i]) List_change(v,lca,dis[u]+dis[v]+edge[i].w);

         if(!mark[i^]) List_change(u,lca,dis[u]+dis[v]+edge[i].w);

     }

     for(int i=;i<=n;i++)

     {

         int tmp=Tree_query(,id[i]);

         if(tmp==INF) puts("-1");

         else printf("%d\n",tmp-dis[i]);

     }

     return ;

 }

唉、