题目链接https://nanti.jisuanke.com/t/28852

题目大意是 h*w 的平面,每两个点有且仅有一条路径,类似于封闭的联通空间,然后在这h*w个点中选取(标记为1~N)N个点(给了坐标),求从1号点按1~N的顺序走到N号点的路程。

练习赛的时候没有思路,队友说可以建树,但还是不清不楚没写出来。

做法是LCA。

将封闭的联通空间建树(点的位置与字符的位置有点麻烦),然后按顺序求两点的最近的公共祖先求深度得距离,最后得路程,算是一道LCA的模板。

 #include <bits/stdc++.h>
#define fir first
#define sec second
#define EPS 1e-12
using namespace std; typedef long long LL;
typedef pair<int , int > pii;
const int MAXN=+;
const int DEG=; struct Edge{
int to,nxt;
}edge[MAXN*];
int head[MAXN],tot; void addEdge(int u,int v){
edge[tot].to=v;
edge[tot].nxt=head[u];
head[u]=tot++;
}
void init(){
tot=;
memset(head,-,sizeof(head));
}
int fa[MAXN][DEG];
int deg[MAXN]; void BFS(int root){
queue< int > que;
deg[root]=;
fa[root][]=root;
que.push(root);
while(!que.empty()){
int tmp=que.front();que.pop();
for(int i=;i<DEG;++i)
fa[tmp][i]=fa[fa[tmp][i-]][i-];
for(int i=head[tmp];i!=-;i=edge[i].nxt){
int v=edge[i].to;
if(v==fa[tmp][]) continue;
deg[v]=deg[tmp]+;
fa[v][]=tmp;
que.push(v);
}
}
} int LCA(int u,int v){
if(deg[u]>deg[v]) swap(u,v);
int hu=deg[u],hv=deg[v];
int tu=u,tv=v;
for(int det=hv-hu,i=;det;det>>=,++i)
if(det&) tv=fa[tv][i];
if(tu==tv) return tu;
for(int i=DEG-;i>=;--i){
if(fa[tu][i]==fa[tv][i])
continue;
tu=fa[tu][i];
tv=fa[tv][i];
}
return fa[tu][];
} char maze[][];
int H,W; void Judge(int cur,int xx,int yy){
int i=xx,j=(yy+)/;
if(maze[xx-][yy]!='_'){
int to=(i-)*W+j;
addEdge(cur,to);
}
if(maze[xx][yy]!='_'){
int to=(i)*W+j;
addEdge(cur,to);
}
if(maze[xx][yy-]!='|'){
int to=(i-)*W+j-;
addEdge(cur,to);
}
if(maze[xx][yy+]!='|'){
int to=(i-)*W+j+;
addEdge(cur,to);
}
} LL caldist(int u,int v){
int ace=LCA(u,v);
return deg[u]+deg[v]-*deg[ace];
} int main()
{
init();
scanf("%d%d%*c",&H,&W);
for(int i=;i<=H;++i){
scanf("%[^\n]%*c",maze[i]);
}
for(int i=;i<=H;++i){
for(int j=;j<=W;++j){
int tmp=(i-)*W+j;
int xx=i,yy=*j-;
Judge(tmp,xx,yy);
}
}
BFS();
LL ans=;
int N,xi,yi;
scanf("%d%d%d",&N,&xi,&yi);
int old=(xi-)*W+yi,aft;
for(int i=;i<N;++i){
scanf("%d%d",&xi,&yi);
aft=(xi-)*W+yi;
ans+=caldist(old,aft);
old=aft;
}
printf("%lld\n",ans);
return ;
}

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