1 /*
2 题意：给出n个岛屿，m座桥连接，每座桥由w个人守住，给出一个炸弹可以炸毁一座桥，而炸每座桥之前
3 要先将桥上的敌人消灭，最少需要w人，问最少需要多少人才能炸毁其中的一座桥
4
5 题解：无向图求桥
6 注意：有重边，当桥上人数为0时，最少应该需要1人(需要有一个人带着炸弹过去)
7 */
8 #include <cstdio>
9 #include <cstring>
10 #include <algorithm>
11
12 using namespace std;
13
14 #define MAXN 1005
15 #define MAXM 1000005
16
17 const int INF = 100000;
18
19 struct node {
20 int v, w, pre, id; //id：边的编号（处理重边情况）
21 } edge[MAXM];
22 int pos[MAXN], nEdge; //图的存储：链式前向星（池子法）
23
24 struct Bridge {
25 int u, v, w;
26 } bridge[MAXM]; //用来记录桥
27 int tot; //桥的个数
28
29 int fa[MAXN], cc; //fa：各个点所属的缩点（连通块），cc连通块的个数
30 int dfn[MAXN], low[MAXN], time; //时间戳
31 int stack[MAXN], top; //用于维护连通块的
32 int n, m; //点的个数和边的条数
33
34 void connect(int u, int v, int w, int id) {
35 nEdge++;
36 edge[nEdge].pre = pos[u];
37 edge[nEdge].v = v;
38 edge[nEdge].w = w;
39 edge[nEdge].id = id;
40 pos[u] = nEdge;
41 }
42
43 void tarjan(int cur, int from) {
44 low[cur] = dfn[cur] = time++;
45 stack[++top] = cur;
46 for (int p=pos[cur]; p; p=edge[p].pre) {
47 int v = edge[p].v;
48 if (edge[p].id == from) continue;
49 if (!dfn[v]) {
50 tarjan(v, edge[p].id);
51 if (low[v] < low[cur]) low[cur] = low[v];
52 if (low[v] > dfn[cur]) {
53 bridge[tot].u = cur;
54 bridge[tot].w = edge[p].w;
55 bridge[tot++].v = v;
56 cc++;
57 do {
58 fa[stack[top]] = cc;
59 } while (stack[top--] != v);
60 }
61 } else if (low[cur] > dfn[v]) low[cur] = dfn[v];
62 }
63 }
64 int main(void)
65 {
66 while(~scanf("%d%d", &n, &m), n || m)
67 {
68 memset(pos, 0, sizeof(pos));
69 nEdge = 0;
70 int u, v, w;
71 for (int i=0; i<m; i++) {
72 scanf("%d%d%d", &u, &v, &w);
73 connect(u, v, w, i);
74 connect(v, u, w, i);
75 }
76
77 memset(dfn, 0, sizeof(dfn));
78 memset(fa, -1, sizeof(fa));
79
80 cc = tot = 0;
81 if (!dfn[1]) {
82 top = time = 1;
83 tarjan(1, -1);
84 ++cc;
85 int j;
86 for (j=1; j<=n; j++) //特殊处理顶点的连通块
87 {
88 if (dfn[j])
89 {
90 if (fa[j] == -1)
91 fa[j] = cc;
92 }
93 else
94 break;
95 }
96 if (j <= n)
97 printf("0\n");
98 else
99 {
100 int minw = INF;
101 if (tot > 0)
102 {
103 for(int i=0; i<tot; i++)
104 minw = min(bridge[i].w,minw);
105 if (minw == 0)
106 minw = 1;
107 printf("%d\n",minw);
108 }
109 else
110 printf("-1\n");
111 }
112 }
113 }
114 return 0;
115 }
