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• # POJ 1470 Closest Common Ancestors(LCA&RMQ)

题意比较费劲：输入看起来很麻烦。处理括号冒号的时候是用%1s就可以。还有就是注意它有根节点。。。Q次查询

在线st算法

```/*************************************************************************
> File Name:            3.cpp
> Author:               Howe_Young
> Mail:                 1013410795@qq.com
> Created Time:         2015年10月08日 星期四 19时03分30秒
************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 10000;
struct Edge {
int to, next;
}edge[maxn<<1];
int ans[maxn];
int Euler[maxn<<1];
int R[maxn];
int dep[maxn];
int dp[maxn<<1][20]; // RMQ
bool in[maxn];
int cnt;
void init()
{
cnt = 0;
tot = 0;
memset(ans, 0, sizeof(ans));
memset(in, false, sizeof(in));
}
{
edge[tot].to = v;
}
void dfs(int u, int depth)
{
Euler[++cnt] = u;
R[u] = cnt;
dep[cnt] = depth;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
dfs(v, depth + 1);
Euler[++cnt] = u;
dep[cnt] = depth;
}
}

void RMQ(int n)
{
for (int i = 1; i <= n; i++) dp[i][0] = i;
int m = (int)(log(n) / log(2));
for (int j = 1; j <= m; j++)
{
for (int i = 1; i + (1 << j) - 1 <= n; i++)
dp[i][j] = dep[dp[i][j - 1]] < dep[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
}
}
int query(int u, int v)
{
int l = R[u], r = R[v];
if (l > r) swap(l, r);
int k = (int)(log(r - l + 1) / log(2));
int lca = dep[dp[l][k]] < dep[dp[r - (1 << k) + 1][k]] ? dp[l][k] : dp[r - (1 << k) + 1][k];
return Euler[lca];
}
int main()
{
int n;
while (~scanf("%d", &n))
{
init();
char s1[3], s2[3];
int u, v, m;
for (int i = 0; i < n; i++)
{
scanf("%d %1s %1s %d %1s", &u, s1, s1, &m, s2);
for (int j = 0; j < m; j++)
{
scanf("%d", &v);
in[v] = true;
}
}
for (int i = 1; i <= n; i++)
if (!in[i])
{
dfs(i, 1);
break;
}
RMQ(cnt);
int Q;
scanf("%d", &Q);
while (Q--)
{
scanf("%1s %d %d %1s", s1, &u, &v, s2);
ans[query(u, v)]++;
}
for (int i = 1; i <= n; i++)
if (ans[i])
printf("%d:%d
", i, ans[i]);
}
return 0;
}```

离线tarjan算法：

```#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000;//节点
const int maxm = 1000010;//最大查询数
struct Edge {
int to, next;
}edge[maxn * 2];
struct Query {
int q, next;
int index;
}query[maxm * 2];
int cnt, h[maxn];
int fa[maxn];
int r[maxn];
int ancestor[maxn];
int ans[maxm];
int Q;
bool vis[maxn];

void init(int n)
{
Q = 0;
tot  = 0;
cnt = 0;
memset(h, -1, sizeof(h));
memset(fa, -1, sizeof(fa));
memset(ancestor, 0, sizeof(ancestor));
memset(vis, false, sizeof(vis));
for (int i = 1; i <= n; i++) r[i] = 1;
}
{
edge[tot].to = v;
}
void addquery(int u, int v, int index)
{
query[cnt].q = v;
query[cnt].index = index;
query[cnt].next = h[u];
h[u] = cnt++;
}
int find(int x)
{
if (fa[x] == -1) return x;
return fa[x] = find(fa[x]);
}
void Union(int x, int y)
{
int tx = find(x);
int ty = find(y);
if (tx != ty)
{
if (tx < ty)
{
fa[tx] = ty;
r[ty] += r[tx];
}
else
{
fa[ty] = tx;
r[tx] += r[ty];
}
}
}
void LCA(int u)
{
vis[u] = true;
ancestor[u] = u;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (vis[v]) continue;
LCA(v);
Union(u, v);
ancestor[find(u)] = u;
}
for (int i = h[u]; i != -1; i = query[i].next)
{
int v = query[i].q;
if (vis[v])
{
ans[query[i].index] = ancestor[find(v)];
}
}
}
int res[maxn];
bool in[maxn];
int main()
{
int n;
while (~scanf("%d", &n))
{
init(n);
memset(in, false, sizeof(in));
int u, v, m;
char ch[2];
for (int i = 1; i <= n; i++)
{
scanf("%d %1s %1s %d %1s", &u, ch, ch, &m, ch);
for (int j = 0; j < m; j++)
{
scanf("%d", &v);
in[v] = true;
}
}
int q;
scanf("%d", &q);
while (q--)
{
scanf("%1s %d %d %1s", ch, &u, &v, ch);
}
int root;
for (int i = 1; i <= n; i++)
{
if (!in[i])
{
root = i;
break;
}
}
LCA(root);
memset(res, 0, sizeof(res));
for (int i = 0; i < Q; i++)
res[ans[i]]++;
for (int i = 1; i <= n; i++)
if (res[i])
printf("%d:%d
", i, res[i]);
}
return 0;
}```
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• 原文地址：https://www.cnblogs.com/Howe-Young/p/4862255.html