Caocao's Bridges HDU - 4738 找桥

 1 //题解：
 2 //这道题。。。。。。T_T
 3 //这道题要先判断有没有桥，没有桥的话就直接输出-1
 4 //如果有桥的话还要注意如果有一座桥上没有警卫，但是我们还是要派一人去炸桥
 5 //所以要有两个特判，卧槽！
 6 #include<stdio.h>
 7 #include<string.h>
 8 #include<iostream>
 9 #include<algorithm>
10 #include<queue>
11 #include<map>
12 #include<vector>
13 using namespace std;
14 const int maxn = 1005;
15 const int INF=0x3f3f3f3f;
16 struct node {
17     int v, next,w;
18 }e[maxn*maxn*2+10];
19 int head[maxn], cnt;
20 bool iscut[maxn];
21 int dfn[maxn], low[maxn];
22 int num,n,m,minn;
23 inline void add_edge(int xx, int yy,int zz) {
24     e[++cnt].next = head[xx];
25     e[cnt].w=zz;
26     e[cnt].v = yy;
27     head[xx] = cnt;
28 }
29 
30 inline void tarjan(int x, int in_edge) {
31     dfn[x] = low[x] = ++num;
32     for(int i = head[x]; i; i = e[i].next) {
33         int to = e[i].v;
34         if(!dfn[to]) {
35             tarjan(to, i);
36             low[x] = min(low[x], low[to]);
37             if(low[to] > dfn[x])
38                 iscut[i] = iscut[i ^ 1] = true,minn=min(minn,e[i].w);
39         }
40         else if(i != (in_edge ^ 1))
41             low[x] = min(low[x], dfn[to]);
42     }
43 }
44 int main()
45 {
46     while(~scanf("%d%d",&n,&m) && (n+m))
47     {
48 
49         cnt=1;
50         num=0;
51         memset(iscut,0,sizeof(iscut));
52         memset(head,0,sizeof(head));
53         memset(dfn,0,sizeof(dfn));
54         memset(low,0,sizeof(low));
55         while(m--)
56         {
57             int x,y,z;
58             scanf("%d%d%d",&x,&y,&z);
59             add_edge(x,y,z);
60             add_edge(y,x,z);
61         }
62         int flag=0;
63         minn=INF;
64         for(int i=1;i<=n;++i)
65         {
66             if(!dfn[i])
67             {
68                 tarjan(i,0);
69                 flag++;
70             }
71         }
72         if(flag>1)
73         {
74             printf("0\n");
75         }
76         else
77         {
78             if(minn==INF) printf("-1\n");
79             else if(minn==0) printf("1\n");
80             else printf("%d\n",minn);
81         }
82     }
83     return 0;
84 }