10765 - Doves and bombs(雙連通分量）

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UVA 10765 - Doves and bombs

題目連結

題意：給定一個無向圖，每個點的鴿子值為刪去這個點後會有幾個連通塊，問鴿子值前m大的點的鴿子值，如果相同，按編號排

思路：就裸的雙連通分量，在每個連通分量如果是割頂的點就加一，最後如果答案為0的點，答案應該是1

代碼：

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>using namespace std;const int N = 10005;struct Edge {int u, v;Edge() {}Edge(int u, int v) {this->u = u;this->v = v;}};int pre[N], bccno[N], dfs_clock, bcc_cnt;bool iscut[N];vector<int> g[N], bcc[N];stack<Edge> S;int dfs_bcc(int u, int fa) {int lowu = pre[u] = ++dfs_clock;int child = 0;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];Edge e = Edge(u, v);if (!pre[v]) {S.push(e);child++;int lowv = dfs_bcc(v, u);lowu = min(lowu, lowv);if (lowv >= pre[u]) {iscut[u] = true;bcc_cnt++; bcc[bcc_cnt].clear(); //start from 1while(1) {Edge x = S.top(); S.pop();if (bccno[x.u] != bcc_cnt) {bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt;}if (bccno[x.v] != bcc_cnt) {bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt;}if (x.u == u && x.v == v) break;}}} else if (pre[v] < pre[u] && v != fa) {S.push(e);lowu = min(lowu, pre[v]);}}if (fa < 0 && child == 1) iscut[u] = false;return lowu;}void find_bcc(int n) {memset(pre, 0, sizeof(pre));memset(iscut, 0, sizeof(iscut));memset(bccno, 0, sizeof(bccno));dfs_clock = bcc_cnt = 0;for (int i = 0; i < n; i++)if (!pre[i]) dfs_bcc(i, -1);}int n, m;struct Node {int val, id;bool operator < (const Node& c) const {if (val != c.val)return val > c.val;return id < c.id;}} node[N];int main() {while (~scanf("%d%d", &n, &m) && n) {int u, v;for (int i = 0; i < n; i++) {g[i].clear();node[i].val = 0;node[i].id = i;}while (scanf("%d%d", &u, &v)) {if (u == -1 && v == -1)break;g[u].push_back(v);g[v].push_back(u);}find_bcc(n);for (int i = 1; i <= bcc_cnt; i++) {for (int j = 0; j < bcc[i].size(); j++) {int u = bcc[i][j];if (iscut[u])node[u].val++;}}sort(node, node + n);for (int i = 0; i < m; i++) {if (node[i].val == 0) node[i].val = 1;printf("%d %d\n", node[i].id, node[i].val);}printf("\n");}return 0;}

10765 - Doves and bombs(雙連通分量）