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
4 5 1 2 2 1 3 2 3 4 4 3 2 1 2 4 3
3 3 6
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
1 #include <algorithm> 2 #include <cstdio> 3 #include <queue> 4 5 using namespace std; 6 7 const int INF(0x3f3f3f3f); 8 const int M(6e5+5); 9 const int N(4e5+5); 10 int n,m; 11 12 int head[N],sumedge=1; 13 struct Edge 14 { 15 int u,v,next,w; 16 Edge(int u=0,int v=0,int next=0,int w=0): 17 u(u),v(v),next(next),w(w){} 18 }edge[M<<1]; 19 inline void ins(int u,int v,int w) 20 { 21 edge[++sumedge]=Edge(u,v,head[u],w); 22 head[u]=sumedge; 23 } 24 25 struct Node 26 { 27 int id,dis; 28 bool operator < (const Node &x) const 29 { 30 return dis>x.dis; 31 } 32 33 }; 34 priority_queue<Node>que; 35 bool vis[N],mark[N]; 36 int dis[N],pre[N]; 37 inline void Dijkstra() 38 { 39 40 for(int i=0;i<=n;i++) dis[i]=INF; 41 dis[1]=0; 42 que.push((Node){1,0}); 43 for(;!que.empty();) 44 { 45 int u=que.top().id; 46 que.pop(); 47 if(vis[u]) continue; 48 vis[u]=1; 49 for(int i=head[u];i;i=edge[i].next) 50 { 51 int v=edge[i].v; 52 if(dis[v]>dis[u]+edge[i].w) 53 { 54 dis[v]=dis[u]+edge[i].w; 55 mark[pre[v]]=0; 56 mark[i]=1; 57 pre[v]=i; 58 que.push((Node){v,dis[v]}); 59 } 60 } 61 } 62 } 63 64 int size[N],deep[N],dad[N],top[N],son[N],cnt,id[N],dfn[N]; 65 void DFS(int x,int fa,int deepth) 66 { 67 size[x]=1; deep[x]=deepth; dad[x]=fa; 68 for(int i=head[x];i;i=edge[i].next) 69 if(mark[i]) 70 { 71 int v=edge[i].v; 72 if(fa==v) continue; 73 DFS(v,x,deepth+1); 74 size[x]+=size[v]; 75 if(size[son[x]]<size[v]) son[x]=v; 76 } 77 } 78 void DFS_(int x,int Top) 79 { 80 id[x]=++cnt; dfn[cnt]=x; 81 top[x]=Top; 82 if(son[x]) DFS_(son[x],Top); 83 for(int i=head[x];i;i=edge[i].next) 84 if(mark[i]) 85 { 86 int v=edge[i].v; 87 if(dad[x]!=v&&son[x]!=v) DFS_(v,v); 88 } 89 } 90 int LCA(int x,int y) 91 { 92 for(;top[x]!=top[y];x=dad[top[x]]) 93 if(deep[top[x]]<deep[top[y]]) swap(x,y); 94 return deep[x]<deep[y]?x:y; 95 } 96 97 struct Tree 98 { 99 int l,r,minn,flag; 100 }tr[N<<2]; 101 #define lc (now<<1) 102 #define rc (now<<1|1) 103 #define mid (tr[now].l+tr[now].r>>1) 104 inline void Tree_up(int now) 105 { 106 tr[now].minn=min(tr[lc].minn,tr[rc].minn); 107 } 108 void Tree_build(int now,int l,int r) 109 { 110 tr[now].l=l; tr[now].r=r; 111 tr[now].minn=INF; 112 tr[now].flag=INF; 113 if(l==r) return ; 114 Tree_build(lc,l,mid); 115 Tree_build(rc,mid+1,r); 116 } 117 inline void Tree_down(int now) 118 { 119 if(tr[now].flag==INF) return ; 120 tr[lc].flag=min(tr[lc].flag,tr[now].flag); 121 tr[rc].flag=min(tr[rc].flag,tr[now].flag); 122 tr[lc].minn=min(tr[lc].minn,tr[lc].flag); 123 tr[rc].minn=min(tr[rc].minn,tr[rc].flag); 124 } 125 void Tree_change(int now,int l,int r,int x) 126 { 127 if(tr[now].l==l&&tr[now].r==r) 128 { 129 tr[now].minn=min(tr[now].minn,x); 130 tr[now].flag=min(tr[now].flag,x); 131 return ; 132 } 133 Tree_down(now); 134 if(r<=mid) Tree_change(lc,l,r,x); 135 else if(l>mid) Tree_change(rc,l,r,x); 136 else Tree_change(lc,l,mid,x),Tree_change(rc,mid+1,r,x); 137 Tree_up(now); 138 } 139 int Tree_query(int now,int to) 140 { 141 if(tr[now].l==tr[now].r) return tr[now].minn; 142 Tree_down(now); 143 if(to<=mid) return Tree_query(lc,to); 144 else return Tree_query(rc,to); 145 } 146 inline void List_change(int u,int v,int w) 147 { 148 int fx=top[u] ; 149 while(deep[fx]>deep[v]) 150 { 151 Tree_change(1,id[top[u]],id[u],w); 152 u=dad[fx],fx=top[u]; 153 } 154 if(u!=v) Tree_change(1,id[v]+1,id[u],w); 155 } 156 157 int main() 158 { 159 freopen("travel.in","r",stdin); 160 freopen("travel.out","w",stdout); 161 162 scanf("%d%d",&n,&m); 163 for(int a,b,t,i=1;i<=m;i++) 164 { 165 scanf("%d%d%d",&a,&b,&t); 166 ins(a,b,t); ins(b,a,t); 167 } 168 Dijkstra(); 169 DFS(1,0,1); 170 DFS_(1,1); 171 Tree_build(1,1,cnt); 172 for(int lca,u,v,i=2;i<=sumedge;i+=2) 173 { 174 u=edge[i].u,v=edge[i].v,lca=LCA(u,v); 175 if(!mark[i]) List_change(v,lca,dis[u]+dis[v]+edge[i].w); 176 if(!mark[i^1]) List_change(u,lca,dis[u]+dis[v]+edge[i].w); 177 } 178 for(int i=2;i<=n;i++) 179 { 180 int tmp=Tree_query(1,id[i]); 181 if(tmp==INF) puts("-1"); 182 else printf("%d ",tmp-dis[i]); 183 } 184 return 0; 185 }
唉、