CodeForces 487E UOJ 30 Tourists Tarjan + 树链剖分

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施工中。。。

以下代码需要C++11才可编译。。

#include <cstdio>#include <algorithm>#include <functional>#include <set>#include <cstring>using namespace std;#define FOR(i,j,k) for(i=j;i<=k;i++)int read() {    int s = 0, f = 1; char ch = getchar();    for (; '0' > ch || ch > '9'; ch = getchar()) if (ch == '-') f = -1;    for (; '0' <= ch && ch <= '9'; ch = getchar()) s = s * 10 + ch - '0';    return s * f;}const int N = 100001, M = 2 * N;struct Graph {    int head[N], next[M], to[M], cnt;    Graph() : cnt(0) {}    void add(int u, int v) {        next[++cnt] = head[u]; head[u] = cnt; to[cnt] = v;        next[++cnt] = head[v]; head[v] = cnt; to[cnt] = u;    }};struct BCC : Graph {    int dfn[N], low[N], id[N], belong[N], cut[N], cnt;    multiset<int> val[M];    BCC() {        cnt = 0;    }    function<void(int,int)> found_cut_point;        void tarjan(int u) {        static int ts = 0, top = 0;        static int stack[N];        int x, i;        dfn[u] = low[u] = ++ts;        stack[++top] = u;        for (i = head[u]; i; i = next[i])            if (!dfn[to[i]]) {                tarjan(to[i]);                low[u] = min(low[u], low[to[i]]);                if (dfn[u] == low[to[i]]) {                    cnt ++;                    cut[u] = 1;                    do {                        x = stack[top--];                        id[x] = cnt;                        if (cut[x])                            found_cut_point(x, cnt);                    } while(x != to[i]);                    found_cut_point(u, cnt);                }            } else low[u] = min(low[u], dfn[to[i]]);    }} bcc;struct SegmentTree {    int initial[M], mi[M];        int build(int t, int l, int r) {        if (l == r) return mi[t] = initial[l];        int mid = l + r >> 1;        return mi[t] = min(build(t * 2, l, mid), build(t * 2 + 1, mid + 1, r));    }        int query(int t, int l, int r, int ql, int qr) {        if (l == ql && r == qr) return mi[t];        int mid = l + r >> 1;        if (qr <= mid) return query(t * 2, l, mid, ql, qr);        else if (ql > mid) return query(t * 2 + 1, mid + 1, r, ql, qr);        else return min(query(t * 2, l, mid, ql, mid),            query(t * 2 + 1, mid + 1, r, mid + 1, qr));    }        void modify(int t, int l, int r, int x, int v) {        if (l == r) { mi[t] = v; return; }         int mid = l + r >> 1;        if (x <= mid) modify(t * 2, l, mid, x, v);        else modify(t * 2 + 1, mid + 1, r, x, v);        mi[t] = min(mi[t * 2], mi[t * 2 + 1]);    }} seg;struct Tree : Graph {    int son[N], sz[N], top[N], dep[N], pos[N], fa[N], id;    SegmentTree *st;    Tree() {        id = 0;    }        void dfs1(int x) {        son[x] = 0; sz[x] = 1;        for (int i = head[x]; i; i = next[i])            if (to[i] != fa[x]) {                fa[to[i]] = x; dep[to[i]] = dep[x] + 1;                dfs1(to[i]);                sz[x] += sz[to[i]];                if (sz[to[i]] > sz[son[x]]) son[x] = to[i];            }    }        void dfs2(int x, int t) {        top[x] = t; pos[x] = ++id;        if (son[x]) dfs2(son[x], t);        for (int i = head[x]; i; i = next[i])            if (to[i] != fa[x] && to[i] != son[x])                dfs2(to[i], to[i]);    }        int lca(int x, int y) {        int fx = top[x], fy = top[y];        while (fx != fy) {            if (dep[fx] < dep[fy]) swap(fx, fy), swap(x, y);            x = fa[fx], fx = top[x];        }        if (dep[x] > dep[y]) swap(x, y);        return x;    }        int query(int x, int y) {        int fx = top[x], fy = top[y], ans = 2147483647;        while (fx != fy) {            if (dep[fx] < dep[fy]) swap(fx, fy), swap(x, y);            ans = min(ans, st->query(1, 1, id, pos[fx], pos[x]));            x = fa[fx], fx = top[x];        }        if (dep[x] > dep[y]) swap(x, y);        return min(ans, st->query(1, 1, id, pos[x], pos[y]));    }} tree;int main() {    static int w[N];    char ch[2];    int n = read(), m = read(), q = read(), i, x, y;    FOR (i, 1, n) w[i] = read();    tree.st = &seg;    bcc.found_cut_point = [n] (int id, int bcc) {        tree.add(id + n, bcc);    };    while (m--) bcc.add(read(), read());    bcc.tarjan(1);    FOR (i, 1, n) {        if (i != 1)            bcc.val[bcc.id[i]].insert(w[i]);        if (bcc.cut[i])            bcc.val[n + i].insert(2147483647);    }    tree.dfs1(n+1); tree.dfs2(n+1, n+1);    memset(seg.initial, 127, sizeof seg.initial);    FOR (i, 1, 2*n) if(tree.pos[i]) seg.initial[tree.pos[i]] = *bcc.val[i].begin();    seg.build(1, 1, tree.id);    while (q--) {        scanf("%s%d%d", ch, &x, &y);        if (ch[0] == 'C') {            if (x != 1) {                bcc.val[bcc.id[x]].erase(bcc.val[bcc.id[x]].find(w[x]));                bcc.val[bcc.id[x]].insert(y);                seg.modify(1, 1, tree.id, tree.pos[bcc.id[x]], *bcc.val[bcc.id[x]].begin());            }            w[x] = y;        } else {            if (x == y) printf("%d\n", w[x]);            else {                x=bcc.cut[x] ? x + n : bcc.id[x];                  y=bcc.cut[y] ? y + n : bcc.id[y];                int l = tree.lca(x, y);                if (l <= bcc.cnt) l = tree.fa[l];                printf("%d\n", min(w[l - n], tree.query(x, y));            }        }    }    return 0;}

There are n cities in Cyberland, numbered from 1 to n, connected by m bidirectional roads. The j-th road connects city ajand bj.

For tourists, souvenirs are sold in every city of Cyberland. In particular, city i sell it at a price of wi.

Now there are q queries for you to handle. There are two types of queries:

"C a w": The price in city a is changed to w.
"A a b": Now a tourist will travel from city a to b. He will choose a route, he also doesn't want to visit a city twice. He will buy souvenirs at the city where the souvenirs are the cheapest (possibly exactly at city a or b). You should output the minimum possible price that he can buy the souvenirs during his travel.

More formally, we can define routes as follow:

A route is a sequence of cities [x1, x2, ..., xk], where k is a certain positive integer.
For any 1 ≤ i < j ≤ k, xi ≠ xj.
For any 1 ≤ i < k, there is a road connecting xi and xi + 1.
The minimum price of the route is min(wx1, wx2, ..., wxk).
The required answer is the minimum value of the minimum prices of all valid routes from a to b.

Input

The first line of input contains three integers n, m, q (1 ≤ n, m, q ≤ 105), separated by a single space.

Next n lines contain integers wi (1 ≤ wi ≤ 109).

Next m lines contain pairs of space-separated integers aj and bj (1 ≤ aj, bj ≤ n, aj ≠ bj).

It is guaranteed that there is at most one road connecting the same pair of cities. There is always at least one valid route between any two cities.

Next q lines each describe a query. The format is "C a w" or "A a b" (1 ≤ a, b ≤ n, 1 ≤ w ≤ 109).

Output

For each query of type "A", output the corresponding answer.

Sample test(s)
input
3 3 31231 22 31 3A 2 3C 1 5A 2 3
output
12
input
7 9 412345671 22 51 52 33 42 45 66 75 7A 2 3A 6 4A 6 7A 3 3
output
2153

Note

For the second sample, an optimal routes are:

From 2 to 3 it is [2, 3].

From 6 to 4 it is [6, 5, 1, 2, 4].

From 6 to 7 it is [6, 5, 7].

From 3 to 3 it is [3].

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