114 lines
3.3 KiB
C++
114 lines
3.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 4e5 + 10, M = N << 2; //因为需要装无向图，所以N=2e5,又因为有新图有旧图，都保存到一个数组中，所以需要再乘以2
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//链式前向星
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int e[M], h1[N], h2[N], idx, w[M], ne[M];
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void add(int h[], int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int f[N][16];
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int depth[N], dist[N];
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void bfs() {
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// 1号点是源点
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depth[1] = 1;
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queue<int> q;
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q.push(1);
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while (q.size()) {
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int u = q.front();
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q.pop();
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for (int i = h2[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!depth[j]) {
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q.push(j);
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depth[j] = depth[u] + 1;
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dist[j] = dist[u] + w[i];
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f[j][0] = u;
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for (int k = 1; k <= 15; k++) f[j][k] = f[f[j][k - 1]][k - 1];
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}
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}
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}
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}
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//最近公共祖先
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int lca(int a, int b) {
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if (depth[a] < depth[b]) swap(a, b);
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for (int k = 15; k >= 0; k--)
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if (depth[f[a][k]] >= depth[b]) a = f[a][k];
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if (a == b) return a;
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for (int k = 15; k >= 0; k--)
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if (f[a][k] != f[b][k]) a = f[a][k], b = f[b][k];
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return f[a][0];
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}
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//边双连通分量
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int dfn[N], low[N], ts, stk[N], top;
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vector<int> dcc[N]; //边双中有哪些原始点
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int id[N], dcc_cnt; //原始点x属于哪个边双连通分量，dcc_cnt指边双连通分量个数
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int is_bridge[M]; //记录哪些边是割边
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void tarjan(int u, int fa) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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for (int i = h1[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!dfn[j]) {
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tarjan(j, i);
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low[u] = min(low[u], low[j]);
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if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
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} else if (i != (fa ^ 1))
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low[u] = min(low[u], dfn[j]);
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}
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if (dfn[u] == low[u]) {
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++dcc_cnt; //边双数量+1
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int x;
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do {
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x = stk[top--];
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id[x] = dcc_cnt; // 记录点与边双关系
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dcc[dcc_cnt].push_back(x); // 记录边双中有哪些点
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} while (x != u);
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}
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}
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int n, m, q;
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int main() {
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memset(h1, -1, sizeof h1); // h1是原图的表头
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memset(h2, -1, sizeof h2); // h2是新生成的缩完点的图表头
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scanf("%d %d", &n, &m);
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//原图
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for (int i = 1; i <= m; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(h1, a, b), add(h1, b, a);
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}
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//用tarjan来缩点，将边双连通分量缩点
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, -1);
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//将新图建出来
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for (int u = 1; u <= n; u++)
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for (int i = h1[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (id[u] != id[j])
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add(h2, id[u], id[j]), add(h2, id[j], id[u]);
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}
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//随便找个存在的号作为根节点，预处理出每个点到根节点的距离
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bfs();
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scanf("%d", &q); // q次询问
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for (int i = 1; i <= q; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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a = id[a], b = id[b];
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printf("%d\n", depth[a] + depth[b] - depth[lca(a, b)] * 2); //这就很显然了
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}
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return 0;
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} |