题目链接
题解:
求桥模板题
需要按照顺序输出桥
代码：
#include #include <string.h> #include #include #include #include using namespace std; const int maxn = 5010;//点数 const int maxm = 20010;//边数，因为是无向图，所以这个值要*2 struct Edge { int to,next; bool cut;//是否是桥标记 } edge[maxm]; int head[maxn],tot; int low[maxn],dfn[maxn],Stack[maxn],belong[maxn];//belong数组的值是1~scc int Index,top; int scc;//边双连通块数/强连通分量的个数 bool Instack[maxn]; int bridge;//桥的数目 bool cut[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; 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]; if(low[v] > dfn[u]) { bridge++; edge[i].cut = true; edge[i^1].cut = true; } if(u == pre && son > 1)cut[u] = true; if(u != pre && low[v] >= dfn[u])cut[u] = true; } else if( Instack[v] && low[u] > dfn[v] ) low[u] = dfn[v]; } if(low[u] == dfn[u]) { scc++; do { v = Stack[--top]; Instack[v] = false; belong[v] = scc; } while( v!=u ); } } void init() { tot = 0; memset(head,-1,sizeof(head)); } void solve(int n) { memset(dfn,0,sizeof(dfn)); memset(Instack,false,sizeof(Instack)); memset(cut,0,sizeof cut); Index = top = scc = 0; bridge = 0; for(int i = 1; i <= n; i++) if(!dfn[i]) Tarjan(i,i); printf("%d critical links\n",bridge); vectorint,int> >ans; for(int u=1; u<=n; u++) { for(int i=head[u]; i!=-1; i=edge[i].next) { if(edge[i].cut && edge[i].to>u)ans.push_back(make_pair(u,edge[i].to));//双向边只取一条边 } } sort(ans.begin(),ans.end()); for(int i=0; i) { printf("%d - %d\n",ans[i].first-1,ans[i].second-1); } printf("\n"); } //处理重边 map<int,int>mapit; inline bool isHash(int u,int v) { if(mapit[u*maxn+v])return true; if(mapit[v*maxn+u])return true; mapit[u*maxn+v] = mapit[v*maxn+u] = 1; return false; } int g[105][105]; int main() { int n; int u,v,m; while(~scanf("%d",&n)) { init(); for(int i=1;i<=n;i++) { scanf("%d (%d)",&u,&m); u++; for(int j=1;j<=m;j++) { scanf("%d",&v); v++; addedge(u,v); addedge(v,u); } } solve(n); } return 0; } /* 5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0 */
