poj 1986 Distance Queries

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LCA

题意:LCA模板题,输入n和m,表示n个点m条边,下面m行是边的信息,两端点和权,后面的那个字母无视掉,没用的。接着k,下面k个询问lca,输出即可

有人说要考虑不连通的情况,我没考虑AC了,另外可能有u,u这样的询问,不过这不影响,照样是写模板,没有特判,一样能过

还是Tarjan快一些

LCA转RMQ在线算法

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define N 40010
#define M 25

int tot;
int __pow[M];
int head[N];
struct edge{
int u,v,w,next;
}e[2*N];
int ver[2*N],R[2*N],first[N],dir[N];
int dp[2*N][25];
bool vis[N];

inline void add(int u , int v ,int w ,int &k)
{
e[k].u = u; e[k].v = v; e[k].w = w;
e[k].next = head[u]; head[u] = k++;
u = u^v; v = u^v; u = u^v;
e[k].u = u; e[k].v = v; e[k].w = w;
e[k].next = head[u]; head[u] = k++;
}

void dfs(int u ,int dep)
{
vis[u] = true; first[u] = ++tot; ver[tot] = u; R[tot] = dep;
for(int k=head[u]; k!=-1; k=e[k].next)
if( !vis[e[k].v] )
{
int v = e[k].v , w = e[k].w;
dir[v] = dir[u] + w;
dfs(v,dep+1);
ver[++tot] = u; R[tot] = dep;
}
}

void ST(int len)
{
int K = (int)(log((double)(len)) / log(2.0));
for(int i=1; i<=len; i++) dp[i][0] = i;
for(int j=1; j<=K; j++)
for(int i=1; i+__pow[j]-1 <= len; i++)
{
int a = dp[i][j-1] , b = dp[i+__pow[j-1]][j-1];
if(R[a] < R[b]) dp[i][j] = a;
else dp[i][j] = b;
}
}

int RMQ(int x ,int y)
{
int K = (int)(log((double)(y-x+1)) / log(2.0));
int a = dp[x][K] , b = dp[y-__pow[K]+1][K];
if(R[a] < R[b]) return a;
else return b;
}

int LCA(int u ,int v)
{
int x = first[u] , y = first[v];
if(x > y) swap(x,y);
int index = RMQ(x,y);
return ver[index];
}

int main()
{
for(int i=0; i<M; i++) __pow[i] = (1<<i);
int n,m,k,str[10];
while(scanf("%d%d",&n,&m)!=EOF)
{
k = 0;
memset(head,-1,sizeof(head));
memset(vis,false,sizeof(vis));
while(m--)
{
int u,v,w;
scanf("%d%d%d%s",&u,&v,&w,str);
add(u,v,w,k);
}
tot = dir[1] = 0;
dfs(1,1);
ST(tot);
int q;
scanf("%d",&q);
while(q--)
{
int u,v,lca;
scanf("%d%d",&u,&v);
lca = LCA(u,v);
printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
}
}
return 0;
}

Tarjan离线算法

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#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 40010
#define M 20010

int head[N];
struct edge{
int u,v,w,next;
}e[2*N];
int __head[N];
struct ask{
int u,v,lca,next;
}ea[M];
int fa[N],ance[N],dir[N];
bool vis[N];

inline void add_edge(int u ,int v ,int w ,int &k)
{
e[k].u = u; e[k].v = v; e[k].w = w;
e[k].next = head[u]; head[u] = k++;
u = u^v; v = u^v; u = u^v;
e[k].u = u; e[k].v = v; e[k].w = w;
e[k].next = head[u]; head[u] = k++;
}

inline void add_ask(int u ,int v ,int &k)
{
ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
ea[k].next = __head[u]; __head[u] = k++;
u = u^v; v = u^v; u = u^v;
ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
ea[k].next = __head[u]; __head[u] = k++;
}

int find(int x){
return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void Tarjan(int u)
{
vis[u] = true;
ance[u] = fa[u] = u;
for(int k=head[u]; k!=-1; k=e[k].next)
if( !vis[e[k].v] )
{
int v = e[k].v , w = e[k].w;
dir[v] = dir[u] + w;
Tarjan(v);
fa[v] = u;
}
for(int k=__head[u]; k!=-1; k=ea[k].next)
if( vis[ea[k].v] )
ea[k].lca = ea[k^1].lca = ance[find(ea[k].v)];
}

int main()
{
int n,m,q,k; char str[10];
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(head,-1,sizeof(head));
memset(__head,-1,sizeof(__head));
memset(vis,false,sizeof(vis));
k = 0;
while(m--)
{
int u,v,w;
scanf("%d%d%d%s",&u,&v,&w,str);
add_edge(u,v,w,k);
}
scanf("%d",&q);
k = 0;
for(int i=0; i<q; i++)
{
int u ,v;
scanf("%d%d",&u,&v);
add_ask(u,v,k);
}
dir[1] = 0;
Tarjan(1);
for(int i=0; i<q; i++)
{
int s = i*2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca;
printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
}
}
return 0;
}