Network UVA - 315 (连通图求割点数)
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> using namespace std; /* * 求 无向图的割点和桥 * 可以找出割点和桥,求删掉每个点后增加的连通块。 * 需要注意重边的处理,可以先用矩阵存,再转邻接表,或者进行判重 */ const int MAXN = 10010; const int MAXM = 100010; struct Edge { int to,next; bool cut;//是否为桥的标记 }edge[MAXM]; int head[MAXN],tot; int Low[MAXN],DFN[MAXN],Stack[MAXN]; int Index,top; bool Instack[MAXN]; bool cut[MAXN]; int add_block[MAXN];//删除一个点后增加的连通块 int bridge; 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; Stack[top++] = u; Instack[u] = true; 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]; //桥 //一条无向边(u,v)是桥,当且仅当(u,v)为树枝边,且满足DFS(u)<Low(v)。 if(Low[v] > DFN[u]) { bridge++; edge[i].cut = true; edge[i^1].cut = true; } //割点 //一个顶点u是割点,当且仅当满足(1)或(2) (1) u为树根,且u有多于一个子树。 //(2) u不为树根,且满足存在(u,v)为树枝边(或称父子边, //即u为v在搜索树中的父亲),使得DFS(u)<=Low(v) if(u != pre && Low[v] >= DFN[u])//不是树根 { cut[u] = true; add_block[u]++; } } else if( Low[u] > DFN[v]) Low[u] = DFN[v]; } //树根,分支数大于1 if(u == pre && son > 1)cut[u] = true; if(u == pre)add_block[u] = son - 1; Instack[u] = false; top--; } void solve(int N) { memset(DFN,0,sizeof(DFN)); memset(Instack,false,sizeof(Instack)); memset(add_block,0,sizeof(add_block)); memset(cut,false,sizeof(cut)); Index = top = 0; bridge = 0; for(int i = 1;i <= N;i++) if(!DFN[i]) Tarjan(i,i); int ans = 0; for(int i = 1;i <= N;i++) if(cut[i]) ans++; printf("%d\n",ans); } void init() { tot = 0; memset(head,-1,sizeof(head)); } int main() { int n; while(scanf("%d",&n), n) { int a,b; char ch; init(); while(scanf("%d",&a),a) { while(scanf("%d%c",&b,&ch)) { addedge(a,b); addedge(b,a); if(ch == '\n') break; } } solve(n); } return 0; }