UOJ 30 CF Round #278Tourists
Posted ljzalc1022
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Tourists
题目大意
(n) 个 (m) 条边的无向图,每个点有点权 (w_i) , (q) 次询问,每次修改第 (a) 个点的点权为 (w) ,或者查询 (a) 到 (b) 所有路径中,最小的点权
数据范围
(1 le n,m,q le 10^5, 1 le w_i le 10^9)
时空限制
2s,256MB
分析
对于这种在无向图上查询路径信息的题,一般利用圆方树将其转为树上问题
对于一个方点,维护点双中最小的点权,但是如果这样,每修改一次就需要修改所有与它相邻的节点。
对此我们有一个经典方法,对于方点维护除了环顶之外的信息,在圆方树中就相当于方点维护所有它儿子的信息,那么我们修改时只需要修改它的父亲,查询的时候如果 (lca) 为方点,那么还需要加上它的父亲的信息(因为父亲也在此点双中)
Code
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <set>
using namespace std;
template<class T> void read(T & x)
{
x = 0; int f = 1, ch = getchar();
while(ch < ‘0‘ || ch > ‘9‘)
{
if(ch == ‘-‘) f = -1;
ch = getchar();
}
while(ch >= ‘0‘ && ch <= ‘9‘)
{
x = x * 10 - ‘0‘ + ch;
ch = getchar();
}
x *= f;
}
#define lson u << 1, l, mid
#define rson u << 1 | 1, mid + 1, r
const int inf = 1000000000;
const int maxn = 100000 + 5;
const int maxm = 100000 + 5;
const int maxe = maxm * 2;
const int maxnode = maxn * 2;
int n, m, q;
int val[maxnode];
struct edge
{
int to, nex;
edge(int to = 0, int nex = 0) : to(to), nex(nex) {}
} g[maxe];
int head[maxn];
int ecnt;
vector<int> adj[maxnode];
inline void addedge(int u, int v)
{
g[ecnt] = edge(v, head[u]), head[u] = ecnt++;
g[ecnt] = edge(u, head[v]), head[v] = ecnt++;
}
inline void adde(int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
int dfc, dfn[maxn], low[maxn];
int top, sta[maxn];
int bcnt;
multiset<int> bccval[maxn];
void tarjan(int u, int fa)
{
dfn[u] = low[u] = ++dfc;
sta[++top] = u;
for(int i = head[u]; ~ i; i = g[i].nex)
{
int v = g[i].to;
if(v != fa)
{
if(!dfn[v])
{
tarjan(v, u);
low[u] = min(low[u], low[v]);
if(low[v] >= dfn[u])
{
int now = n + (++bcnt);
adde(now, u);
while(true)
{
int x = sta[top--];
bccval[bcnt].insert(val[x]);
adde(now, x);
if(x == v) break;
}
val[now] = * bccval[bcnt].begin();
}
}
else low[u] = min(low[u], dfn[v]);
}
}
}
struct segment_tree
{
int mn[maxnode << 2];
inline void pushup(int u)
{
mn[u] = min(mn[u << 1], mn[u << 1 | 1]);
}
void build(int u, int l, int r)
{
if(l == r)
{
mn[u] = val[ver[l]];
return;
}
int mid = (l + r) >> 1;
build(lson), build(rson);
pushup(u);
}
void update(int u, int l, int r, int qp)
{
if(l == r)
{
mn[u] = val[ver[l]];
return;
}
int mid = (l + r) >> 1;
if(qp <= mid) update(lson, qp);
else update(rson, qp);
pushup(u);
}
int query(int u, int l, int r, int ql, int qr)
{
if(l == ql && r == qr) return mn[u];
int mid = (l + r) >> 1;
if(qr <= mid) return query(lson, ql, qr);
else if(ql > mid) return query(rson, ql, qr);
else
{
int lv = query(lson, ql, mid);
int rv = query(rson, mid + 1, qr);
return min(lv, rv);
}
}
int dep[maxnode], anc[maxnode], siz[maxnode], son[maxnode];
int dfn[maxnode], top[maxnode], ver[maxnode], dfc;
void dfs1(int u)
{
siz[u] = 1;
for(unsigned int i = 0; i < adj[u].size(); ++i)
{
int v = adj[u][i];
if(v != anc[u])
{
dep[v] = dep[u] + 1;
anc[v] = u;
dfs1(v);
siz[u] += siz[v];
if(siz[son[u]] <= siz[v]) son[u] = v;
}
}
}
void dfs2(int u, int chain)
{
top[u] = chain;
dfn[u] = ++dfc;
ver[dfc] = u;
if(son[u])
{
dfs2(son[u], chain);
}
for(unsigned int i = 0; i < adj[u].size(); ++i)
{
int v = adj[u][i];
if(v != anc[u] && v != son[u])
{
dfs2(v, v);
}
}
}
void init()
{
dfs1(1);
dfs2(1, 1);
build(1, 1, n + bcnt);
}
void update(int u, int x)
{
int v = anc[u];
if(v) bccval[v - n].erase(bccval[v - n].find(val[u]));
val[u] = x;
if(v) bccval[v - n].insert(val[u]);
if(v) val[v] = * bccval[v - n].begin();
update(1, 1, n + bcnt, dfn[u]);
if(v) update(1, 1, n + bcnt, dfn[v]);
}
int query(int u, int v)
{
int re = inf;
while(top[u] != top[v])
{
if(dep[top[u]] > dep[top[v]]) swap(u, v);
re = min(re, query(1, 1, n + bcnt, dfn[top[v]], dfn[v]));
v = anc[top[v]];
}
if(dep[u] > dep[v]) swap(u, v);
re = min(re, query(1, 1, n + bcnt, dfn[u], dfn[v]));
if(u > n)
{
re = min(re, val[anc[u]]);
}
return re;
}
} seg;
void init()
{
tarjan(1, 0);
seg.init();
}
int main()
{
// freopen("testdata.in", "r", stdin);
read(n), read(m), read(q);
for(int i = 1; i <= n; ++i)
{
read(val[i]);
}
memset(head, -1, sizeof(head));
for(int i = 1; i <= m; ++i)
{
int u, v; read(u), read(v);
addedge(u, v);
}
init();
for(int i = 1; i <= q; ++i)
{
char op[5]; scanf("%s", op);
if(op[0] == ‘C‘)
{
int a, w; read(a), read(w);
seg.update(a, w);
}
else
{
int a, b; read(a), read(b);
printf("%d
", seg.query(a, b));
}
}
return 0;
}
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