题目链接：https://vjudge.net/problem/HDU-4612
题目：一个大地图，给定若干个连通图，每个连通图中有若干个桥，你可以在任意某个连通图的
任意两个点添加一条边，问，添加一条边后，大地图中最少剩下几个桥。
思路：tarjan缩点，重构图，对每个新图跑两次dfs求出树的直径，取所有新图的直径max，
答案就是 大地图总桥数 - max(树的直径)。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <queue> 5 #include <vector> 6 using namespace std; 7 #define pb push_back 8 9 const int N = (int)2e5+10; 10 const int M = (int)1e6+10; 11 int n,m,bridge,tot,tim,top,scc; 12 int head[N],dfn[N],low[N],s[N],scc_no[N],vis[N]; 13 struct node{ 14 int to; 15 int nxt; 16 }e[M << 1]; 17 vector<int> g[N];//新图 18 vector<int> poi;//存单个连通图包含的点 19 20 void init(){ 21 for(int i = 0; i <= n; ++i){ 22 head[i] = -1; 23 dfn[i] = 0; 24 g[i].clear(); 25 } 26 bridge = tot = tim = top = scc = 0; 27 } 28 29 inline void add(int u,int v){ 30 e[tot].to = v; 31 e[tot].nxt = head[u]; 32 head[u] = tot++; 33 } 34 35 //tarjan缩点 36 void tarjan(int now,int pre){ 37 poi.pb(now);//存下这个连通图包含的点 38 dfn[now] = low[now] = ++tim; 39 s[top++] = now; 40 int to,pre_cnt = 0; 41 for(int o = head[now]; ~o; o = e[o].nxt){ 42 to = e[o].to; 43 if(to == pre && pre_cnt == 0) { pre_cnt = 1; continue; } 44 if(!dfn[to]){ 45 tarjan(to,now); 46 low[now] = min(low[now],low[to]); 47 if(dfn[now] < low[to]) ++bridge; 48 }else low[now] = min(low[now],dfn[to]); 49 } 50 51 if(dfn[now] == low[now]){ 52 int x; 53 ++scc; 54 do{ 55 x = s[--top]; 56 scc_no[x] = scc; 57 }while(now != x); 58 } 59 } 60 61 //对poi中的那些点新建一个图 62 void rebuild(){ 63 int to,now; 64 for(int i = 0; i < (int)poi.size(); ++i){ 65 now = poi[i]; 66 for(int o = head[now]; ~o; o = e[o].nxt){ 67 to = e[o].to; 68 if(scc_no[now] == scc_no[to]) continue; 69 g[scc_no[now]].pb(scc_no[to]); g[scc_no[to]].pb(scc_no[now]); 70 } 71 } 72 } 73 74 void dfs(int now,int pre,int deep){ 75 vis[now] = deep; 76 int to; 77 for(int i = 0; i < (int)g[now].size(); ++i){ 78 to = g[now][i]; 79 if(to == pre || to == now || vis[to]) continue; 80 dfs(to,now,deep+1); 81 } 82 } 83 84 void test01(){ 85 for(int i = 1; i <= n; ++i) 86 printf("%d 属于scc = %d ",i,scc_no[i]); 87 } 88 89 90 void solve(){ 91 92 int max_deep = 0,_deep = 0; 93 int u; 94 for(int i = 1; i <= n; ++i){ 95 if(!dfn[i]){ 96 poi.clear();//清空上个连通图中的点 97 tarjan(i,i); 98 rebuild();//poi中的点重建图 99 dfs(poi[0],poi[0],1); 100 _deep = 0; 101 for(int i = 0; i < poi.size(); ++i){ 102 if(_deep < vis[poi[i]]){ _deep = vis[poi[i]]; u = poi[i]; } 103 vis[poi[i]] = 0; //这里别忘了初始化 不然下次dfs会出错 104 } 105 dfs(u,u,1); 106 for(int i = 0; i < poi.size(); ++i){ 107 _deep = max(_deep,vis[poi[i]]); 108 vis[poi[i]] = 0;//这里别忘了初始化 不然下组数据的dfs会出错 109 } 110 max_deep = max(max_deep,_deep);//对每个连通图跑两次dfs求树的直径，取max 111 } 112 } 113 // cout << bridge << " " << max_deep << endl; 114 printf("%d ",bridge - max_deep +1); 115 } 116 117 int main(){ 118 119 int u,v; 120 while(~scanf("%d%d",&n,&m) && (n+m)){ 121 init(); 122 for(int i = 0; i < m; ++i){ 123 scanf("%d%d",&u,&v); 124 add(u,v); add(v,u); 125 } 126 solve(); 127 } 128 129 130 return 0; 131 }
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