POJ 3177 Redundant Paths / 边双连通分量

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和上一题一样给你一张图求最少加几条边可以使全图双连通多了重边

多加了一个bool的数组判断重边重边只算一条

还是用了有向图的强连通分量与有向图相比多了 1.参数fa 父节点 2.如果子节点是其父节点 continue

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int maxn = 5010;vector <int> G[maxn];bool ok[maxn][maxn];int pre[maxn];int low[maxn];int sccno[maxn];int dfs_clock;int scc_cnt;stack <int> S;int n, m;int degree[maxn];int Topo[maxn][maxn];void dfs(int u, int fa){pre[u] = low[u] = ++dfs_clock;S.push(u);for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(v == fa)continue;if(!pre[v]){dfs(v, u);low[u] = min(low[u], low[v]);}elselow[u] = min(low[u], pre[v]);}if(low[u] == pre[u]){scc_cnt++;while(1){int x = S.top();S.pop();sccno[x] = scc_cnt;if(x == u)break;}}}void find_scc(){dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for(int i = 1; i <= n; i++)if(!pre[i])dfs(i, -1);}int main(){while(scanf("%d %d", &n, &m) != EOF){for(int i = 1; i <= n; i++)G[i].clear();memset(ok, false, sizeof(ok));while(m--){int u, v;scanf("%d %d", &u, &v);if(!ok[u][v]){G[u].push_back(v);G[v].push_back(u);ok[u][v] = true;}}find_scc();memset(degree, 0, sizeof(degree));for(int i = 1; i <= n; i++){for(int j = 0; j < G[i].size(); j++){int v = G[i][j];if(sccno[i] != sccno[v]){degree[sccno[i]]++;degree[sccno[v]]++;}}}int ans = 0;for(int i = 1; i <= scc_cnt; i++)if(degree[i] == 2)ans++;printf("%d\n", (ans + 1) / 2);}return 0;}

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