HDU 4005 The war(双连通好题）

HDU 4005 The war

题目链接

题意：给一个连通的无向图，每条边有一个炸掉的代价，现在要建一条边（你不不知道的），然后你要求一个你需要的最少代价，保证不管他建在哪，你都能炸掉使得图不连通

思路：炸肯定要炸桥，所以先双连通缩点，得到一棵树，树边是要炸的，那么找一个最小值的边，从该边的两点出发，走的路径中，把两条包含最小值的路径，的两点连边，形成一个环，这个环就保证了最低代价在里面，除了这个环以外的最小边，就是答案，这样的话，就利用一个dfs，搜到每个子树的时候进行一个维护即可

代码：

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int N = 10005;const int M = 200005;int n, m;struct Edge {int u, v, val, id;bool iscut;Edge() {}Edge(int u, int v, int val, int id) {this->u = u;this->v = v;this->val = val;this->id = id;this->iscut = false;}} edge[M];int en, first[N], next[M];void init() {en = 0;memset(first, -1, sizeof(first));}void add_edge(int u, int v, int val, int id) {edge[en] = Edge(u, v, val, id);next[en] = first[u];first[u] = en++;}int pre[N], dfn[N], dfs_clock, bccn, bccno[N];vector<Edge> bcc[N];void dfs_cut(int u, int id) {pre[u] = dfn[u] = ++dfs_clock;for (int i = first[u]; i + 1; i = next[i]) {if (edge[i].id == id) continue;int v = edge[i].v;if (!pre[v]) {dfs_cut(v, edge[i].id);dfn[u] = min(dfn[u], dfn[v]);if (dfn[v] > pre[u])edge[i].iscut = edge[i^1].iscut = true;} else dfn[u] = min(dfn[u], pre[v]);}}void find_cut() {dfs_clock = 0;memset(pre, 0, sizeof(pre));for (int i = 1; i <= n; i++)if (!pre[i]) dfs_cut(i, -1);}void dfs_bcc(int u) {bccno[u] = bccn;for (int i = first[u]; i + 1; i = next[i]) {if (edge[i].iscut) continue;int v = edge[i].v;if (bccno[v]) continue;dfs_bcc(v);}}const int INF = 0x3f3f3f3f;Edge Mine;void find_bcc() {bccn = 0;memset(bccno, 0, sizeof(bccno));for (int i = 1; i <= n; i++) {if (!bccno[i]) {bccn++;dfs_bcc(i);}}for (int i = 1; i <= bccn; i++) bcc[i].clear();Mine.val = INF;for (int i = 0; i < en; i++) {if (!edge[i].iscut) continue;if (Mine.val > edge[i].val)Mine = edge[i];int u = bccno[edge[i].u], v = bccno[edge[i].v], w = edge[i].val;bcc[u].push_back(Edge(u, v, w, 0));}}int ans;int dfs(int u, int f) {int Min1 = INF, Min2 = INF;for (int i = 0; i < bcc[u].size(); i++) {int v = bcc[u][i].v;if (v == f) continue;Min2 = min(min(dfs(v, u), bcc[u][i].val), Min2);if (Min2 < Min1) swap(Min1, Min2);}ans = min(ans, Min2);return Min1;}int main() {while (~scanf("%d%d", &n, &m)) {init();int u, v, w;for (int i = 0; i < m; i++) {scanf("%d%d%d", &u, &v, &w);if (u > n || v > n) continue;add_edge(u, v, w, i);add_edge(v, u, w, i);}find_cut();find_bcc();if (bccn == 1) {printf("-1\n");continue;}ans = INF;u = bccno[Mine.u]; v = bccno[Mine.v];dfs(u, v);dfs(v, u);if (ans == INF) ans = -1;printf("%d\n", ans);}return 0;}