标签:style http color io os ar for sp on
题意: 求出删除一个点之后,连通块最多有多少
思路:数组记录每个点删除后的连通块有多少个,注意图不一定是连通的。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <utility> #include <algorithm> using namespace std; const int MAXN = 10005; struct Edge{ int to, next; bool cut; }edge[MAXN * 10]; int head[MAXN], tot; int Low[MAXN], DFN[MAXN], Stack[MAXN]; int Index, cnt; bool Instack[MAXN]; bool cut[MAXN]; int add_block[MAXN]; void addedge(int u, int v) { edge[tot].to = v; edge[tot].next = head[u]; edge[tot].cut = false; head[u] = tot++; } void Tarjan(int u, int pre) { int v; Low[u] = DFN[u] = ++Index; int son = 0; for (int i = head[u]; i != -1; i = edge[i].next) { v = edge[i].to; if (v == pre) continue; if (!DFN[v]) { son++; Tarjan(v, u); if (Low[u] > Low[v]) Low[u] = Low[v]; if (u != pre && Low[v] >= DFN[u]) { cut[u] = true; add_block[u]++; } } else if (Low[u] > DFN[v]) Low[u] = DFN[v]; } if (u == pre && son > 1) cut[u] = true; if (u == pre) add_block[u] = son - 1; } void init() { memset(head, -1, sizeof(head)); memset(DFN, 0, sizeof(DFN)); memset(add_block, 0, sizeof(add_block)); memset(cut, false, sizeof(cut)); tot = Index = cnt = 0; } void solve(int N) { for (int i = 1; i <= N; i++) if (!DFN[i]) { Tarjan(i, i); cnt++; } int ans = 0; for (int i = 1; i <= N; i++) ans = max(ans, cnt + add_block[i]); printf("%d\n", ans); } int main() { int n, m; while (scanf("%d%d", &n, &m) == 2) { if (n == 0 && m == 0) break; init(); int u, v; for (int i = 0; i < m; i++) { scanf("%d%d", &u, &v); u++; v++; addedge(u, v); addedge(v, u); } solve(n); } return 0; }
标签:style http color io os ar for sp on
原文地址:http://blog.csdn.net/u011345461/article/details/40047849