HDU-4738 Caocao's Bridges 边联通分量

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4738
　　题意：在有重边的无向图中，求权值最小的桥。
　　注意trick就好了，ans为0时输出1，总要有一个人去丢炸弹吧。。。
  1 //STATUS:C++_AC_62MS_8144KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 #pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=1010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1000000007,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-6;
 41 const double OO=1e60;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 /* Edge-Biconnected Component(可以处理重边)
 59   iscut[]为割边集
 60   bccno[]为双连通点集,保存为编号        */
 61 struct Edge{
 62     int u,v,w;
 63 }e[2*N*N];
 64 bool iscut[2*N*N];
 65 int first[N],next[2*N*N],pre[N],low[N],bccno[N];
 66 int n,m,mt,bcnt,dfs_clock;
 67 stack<int> s;
 68 
 69 void adde(int a,int b,int c)
 70 {
 71     e[mt].u=a;e[mt].v=b;e[mt].w=c;
 72     next[mt]=first[a];first[a]=mt++;
 73     e[mt].u=b;e[mt].v=a;e[mt].w=c;
 74     next[mt]=first[b];first[b]=mt++;
 75 }
 76 
 77 void dfs(int u,int fa)
 78 {
 79     int i,v;
 80     pre[u]=low[u]=++dfs_clock;
 81     s.push(u);
 82     int cnt=0;
 83     for(i=first[u];i!=-1;i=next[i]){
 84         v=e[i].v;
 85         if(!pre[v]){
 86             dfs(v,u);
 87             low[u]=Min(low[u],low[v]);
 88             if(low[v]>pre[u])iscut[i]=true;   //存在割边
 89         }
 90         else if(fa==v){  //反向边更新
 91             if(cnt)low[u]=Min(low[u],pre[v]);
 92             cnt++;
 93         }
 94         else low[u]=Min(low[u],pre[v]);
 95     }
 96     if(low[u]==pre[u]){  //充分必要条件
 97         int x=-1;
 98         bcnt++;
 99         while(x!=u){
100             x=s.top();s.pop();
101             bccno[x]=bcnt;
102         }
103     }
104 }
105 
106 int find_bcc()
107 {
108     int i,cnt=0;
109     bcnt=dfs_clock=0;
110     mem(pre,0);mem(bccno,0);mem(iscut,0);
111     for(i=1;i<=n;i++){
112         if(!pre[i]){cnt++;dfs(i,-1);}
113     }
114     return cnt;
115 }
116 
117 int main(){
118  //   freopen("in.txt","r",stdin);
119     int i,j,a,b,c,ans,t;
120     while(~scanf("%d%d",&n,&m) && (n||m))
121     {
122         mem(first,-1);mt=0;
123         for(i=0;i<m;i++){
124             scanf("%d%d%d",&a,&b,&c);
125             adde(a,b,c);
126         }
127 
128         t=find_bcc();
129         if(t>1){
130             printf("0
");
131             continue;
132         }
133 
134         ans=INF;
135         for(i=0;i<mt;i++){
136             if(iscut[i])ans=Min(ans,e[i].w);
137         }
138         printf("%d
",ans==INF?-1:(ans?ans:1));
139     }
140     return 0;
141 }