JZOJ3225[BJOI2013]load（圆方树+树链剖分/树上差分）
题目链接
题目大意
给出一张\(N\)个点\(M\)条边的无向图，和\(Q\)对点对\(p_i, q_i\)，问最后图中每个点必定被\(p_i\)到\(q_i\)的路径覆盖多少次
\(N \le 1e5, M, Q \le 2e5\)
| 样例输入 | 样例输出 |
|---|---|
| 4 4 2 1 2 1 3 2 3 1 4 4 2 4 3 |
2 1 1 2 |
解析
考虑\(p\)到\(q\)的路径必定经过哪些点，显然是\(p\)到\(q\)的简单路径经过的割点
回忆圆方树的构造，树上度数大于\(1\)的圆点就是原图上的割点
那么问题就简单了，建出圆方树，然后\(p\)到\(q\)路径上的圆点答案都会加\(1\)（因为题目说起点和终点也算，就不用特殊处理了）
我拿了个树剖来维护，也可以树上差分
代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#define MAXN 100005
typedef long long LL;
struct Graph {
struct Edge {
int v, next;
Edge(int _v = 0, int _n = 0):v(_v), next(_n) {}
} edge[MAXN << 2];
int head[MAXN << 1], cnt;
void init() { memset(head, -1, sizeof head); cnt = 0; }
void add_edge(int u, int v) { edge[cnt] = Edge(v, head[u]); head[u] = cnt++; }
void insert(int u, int v) { add_edge(u, v); add_edge(v, u); }
};
struct SegmentTree {
int add[MAXN << 3];
void update(int, int, int, int, int);
int query(int, int, int, int);
};
char gc();
int read();
void Tarjan(int, int);
void rebuild();
void dfs1(int);
void dfs2(int);
int N, M, Q, idx, tot;
int dfn[MAXN << 1], top[MAXN << 1], fa[MAXN << 1], dep[MAXN << 1], low[MAXN], size[MAXN << 1], heavy[MAXN << 1];
int stk[MAXN], stop;
std::vector<int> bel[MAXN];
Graph G;
SegmentTree sgt;
int main() {
G.init();
tot = N = read(), M = read(), Q = read();
for (int i = 1; i <= M; ++i) G.insert(read(), read());
Tarjan(1, 0);
rebuild();
dfs1(1);
top[1] = 1, idx = 0;
dfs2(1);
while (Q--) {
int p = read(), q = read();
while (top[p] ^ top[q]) {
if (dep[top[p]] < dep[top[q]]) std::swap(p, q);
sgt.update(1, 1, tot, dfn[top[p]], dfn[p]);
p = fa[top[p]];
}
if (dep[p] > dep[q]) std::swap(p, q);
sgt.update(1, 1, tot, dfn[p], dfn[q]);
}
for (int i = 1; i <= N; ++i)
printf("%d\n", sgt.query(1, 1, tot, dfn[i]));
return 0;
}
inline char gc() {
static char buf[1000000], *p1, *p2;
if (p1 == p2) p1 = (p2 = buf) + fread(buf, 1, 1000000, stdin);
return p1 == p2 ? EOF : *p2++;
}
inline int read() {
int res = 0; char ch = gc();
while (ch < '0' || ch > '9') ch = gc();
while (ch >= '0' && ch <= '9') res = (res << 1) + (res << 3) + ch - '0', ch = gc();
return res;
}
void SegmentTree::update(int rt, int L, int R, int l, int r) {
if (L >= l && R <= r) ++add[rt];
else {
int mid = (L + R) >> 1;
if (l <= mid) update(rt << 1, L, mid, l, r);
if (r > mid) update(rt << 1 | 1, mid + 1, R, l, r);
}
}
int SegmentTree::query(int rt, int L, int R, int pos) {
if (L == R) return add[rt];
int mid = (L + R) >> 1;
if (pos <= mid) return add[rt] + query(rt << 1, L, mid, pos);
else return add[rt] + query(rt << 1 | 1, mid + 1, R, pos);
}
void Tarjan(int u, int fa) {
dfn[u] = low[u] = ++idx;
for (int i = G.head[u]; ~i; i = G.edge[i].next) {
int v = G.edge[i].v;
if (v == fa) continue;
if (!dfn[v]) {
stk[stop++] = v;
Tarjan(v, u);
low[u] = std::min(low[u], low[v]);
if (low[v] >= dfn[u]) {
int p; ++tot;
do {
p = stk[--stop];
bel[p].push_back(tot);
} while (p ^ v);
bel[u].push_back(tot);
}
} else low[u] = std::min(low[u], dfn[v]);
}
}
void rebuild() {
G.init();
for (int i = 1; i <= N; ++i) for (int j = 0; j < bel[i].size(); ++j)
G.insert(i, bel[i][j]);
}
void dfs1(int u) {
dep[u] = dep[fa[u]] + 1;
size[u] = 1;
for (int i = G.head[u]; ~i; i = G.edge[i].next) {
int v = G.edge[i].v;
if (v == fa[u]) continue;
fa[v] = u, dfs1(v);
size[u] += size[v];
if (!heavy[u] || size[heavy[u]] < size[v]) heavy[u] = v;
}
}
void dfs2(int u) {
dfn[u] = ++idx;
if (heavy[u]) {
top[heavy[u]] = top[u];
dfs2(heavy[u]);
}
for (int i = G.head[u]; ~i; i = G.edge[i].next) {
int v = G.edge[i].v;
if (v == fa[u] || v == heavy[u]) continue;
top[v] = v, dfs2(v);
}
}
//Rhein_E
