// tarjan算法求無向圖的橋、邊雙連通分量并縮點 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; const int SIZE = 100010; int head[SIZE], ver[SIZE * 2], Next[SIZE * 2]; int dfn[SIZE], low[SIZE], c[SIZE]; int n, m, tot, num, dcc, tc; bool bridge[SIZE * 2]; int hc[SIZE], vc[SIZE * 2], nc[SIZE * 2];void add(int x, int y) {ver[++tot] = y, Next[tot] = head[x], head[x] = tot; }void add_c(int x, int y) {vc[++tc] = y, nc[tc] = hc[x], hc[x] = tc; }void tarjan(int x, int in_edge) {dfn[x] = low[x] = ++num;for (int i = head[x]; i; i = Next[i]) {int y = ver[i];if (!dfn[y]) {tarjan(y, i);low[x] = min(low[x], low[y]);if (low[y] > dfn[x])bridge[i] = bridge[i ^ 1] = true;}else if (i != (in_edge ^ 1))low[x] = min(low[x], dfn[y]);} }void dfs(int x) {c[x] = dcc;for (int i = head[x]; i; i = Next[i]) {int y = ver[i];if (c[y] || bridge[i]) continue;dfs(y);} }int main() {cin >> n >> m;tot = 1;for (int i = 1; i <= m; i++) {int x, y;scanf("%d%d", &x, &y);add(x, y), add(y, x);}for (int i = 1; i <= n; i++)if (!dfn[i]) tarjan(i, 0);for (int i = 2; i < tot; i += 2)if (bridge[i])printf("%d %d\n", ver[i ^ 1], ver[i]);for (int i = 1; i <= n; i++)if (!c[i]) {++dcc;dfs(i);}printf("There are %d e-DCCs.\n", dcc);for (int i = 1; i <= n; i++)printf("%d belongs to DCC %d.\n", i, c[i]);tc = 1;for (int i = 2; i <= tot; i++) {int x = ver[i ^ 1], y = ver[i];if (c[x] == c[y]) continue;add_c(c[x], c[y]);}printf("縮點之后的森林，點數 %d，邊數 %d\n", dcc, tc / 2);for (int i = 2; i < tc; i += 2)printf("%d %d\n", vc[i ^ 1], vc[i]); }

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