132 lines
3.2 KiB
C++
132 lines
3.2 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 3010, M = N << 4;
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const LL mod = 19491001;
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(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 n, m;
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// 最简并查集
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int p[N];
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int find(int x) {
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if (x == p[x]) return x;
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return p[x] = find(p[x]);
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}
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LL f[N];
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int dfn[N], low[N], ts;
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int stk[N], top;
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int bcnt;
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/*
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// 边双模板
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vector<int> bcc[N];
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int dcc[N];
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void tarjan(int u, int from) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (!dfn[v]) {
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tarjan(v, i);
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low[u] = min(low[u], low[v]);
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} else if (i != (from ^ 1))
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low[u] = min(low[u], dfn[v]);
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}
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if (dfn[u] == low[u]) {
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++bcnt;
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int x;
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do {
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x = stk[top--];
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bcc[bcnt].push_back(x); // 记录边双中有哪些点
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dcc[x] = bcnt; // 反向记录x这个点在bcnt这个边双里面
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} while (x != u);
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}
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}
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*/
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int dcc[N]; // 记录每一个节点在哪个边双中
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bool vis[M << 1];
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void tarjan(int u, int fa) {
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low[u] = dfn[u] = ++ts;
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stk[++top] = u;
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for (int i = h[u]; ~i; i = ne[i]) {
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if (vis[i]) continue;
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if ((i == fa) || ((i ^ 1) == fa)) continue;
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int v = e[i];
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if (!dfn[v]) {
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vis[i] = vis[i ^ 1] = 1;
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tarjan(v, fa);
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vis[i] = vis[i ^ 1] = 0;
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low[u] = min(low[u], low[v]);
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} else
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low[u] = min(low[u], dfn[v]);
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}
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if (low[u] == dfn[u]) {
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dcc[u] = ++bcnt;
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while (stk[top] != u)
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dcc[stk[top--]] = bcnt;
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top--;
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}
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}
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int calc(int u, int v) {
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int x = find(u), y = find(v);
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if (x != y) return 0;
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if (dcc[u] != dcc[v]) return 1;
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if (f[u] == f[v]) return 3;
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return 2;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("P4214.in", "r", stdin);
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#endif
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m); // 节点数和管道数
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for (int i = 1; i <= n; i++) p[i] = i; // 初始化并查集
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for (int i = 1; i <= m; i++) { // 枚举每条边
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int a, b;
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scanf("%d %d", &a, &b);
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add(a, b), add(b, a);
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// 并查集这是在干什么？记录点与点的连通性？
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int x = find(a), y = find(b);
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if (x != y) p[x] = y;
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}
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// base case初始化
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for (int i = 1; i <= n; i++) f[i] = 1ll;
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// 为什么是m+1次呢？
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for (int x = 1; x <= m + 1; x++) {
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// tarjan初始化
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memset(low, 0, sizeof low);
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memset(dfn, 0, sizeof dfn);
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ts = 0, bcnt = 0, top = 0;
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// 跑Tarjan,可能不连通
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, x << 1);
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for (int i = 1; i <= n; i++)
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f[i] = f[i] * mod + (LL)dcc[i];
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}
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// 根据题意，计算最终的结果
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int ans = 0;
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for (int i = 1; i <= n; i++)
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for (int j = i + 1; j <= n; j++)
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ans += calc(i, j);
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// 输出
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printf("%d\n", ans);
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return 0;
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} |