HDU 4738 Caocao's Bridges

Caocao's Bridges

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4738
64-bit integer IO format: %I64d Java class name: Main

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and based on those islands, Caocao's army could easily attack Zhou Yu's troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao's army could be deployed very conveniently among those islands. Zhou Yu couldn't stand with that, so he wanted to destroy some Caocao's bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn't be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete the island seperating mission.

Input

There are no more than 12 test cases.

In each test case:

The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N2 )

Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )

The input ends with N = 0 and M = 0.

Output

For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn't succeed any way, print -1 instead.

Sample Input

Sample Output

-1
4

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

解题：注意存在不连通情况。。。草草的桥。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define INF 0x3f3f3f3f
15 #define pii pair<int,int>
16 using namespace std;
17 const int maxn = 1005;
18 struct arc {
19     int to,id,next;
20     arc(int x = 0,int y = 0,int z = -1) {
21         to = x;
22         id = y;
23         next = z;
24     }
25 };
26 arc e[maxn*maxn*2];
27 int head[maxn],dfn[maxn],low[maxn],uf[maxn];
28 int tot,idx,ans,p[maxn*maxn*2],n,m;
29 int Find(int x){
30     if(x == uf[x]) return uf[x];
31     return uf[x] = Find(uf[x]);
32 }
33 void add(int u,int v,int id) {
34     e[tot] = arc(v,id,head[u]);
35     head[u] = tot++;
36     e[tot] = arc(u,id,head[v]);
37     head[v] = tot++;
38 }
39 void tarjan(int u,int fa) {
40     dfn[u] = low[u] = ++idx;
41     bool flag = true;
42     for(int i = head[u]; ~i; i = e[i].next) {
43         if(e[i].to == fa && flag) {
44             flag = false;
45             continue;
46         }
47         if(!dfn[e[i].to]) {
48             tarjan(e[i].to,u);
49             low[u] = min(low[u],low[e[i].to]);
50             if(low[e[i].to] > dfn[u] && p[e[i].id] < ans)
51                 ans = p[e[i].id];
52         } else low[u] = min(low[u],dfn[e[i].to]);
53     }
54 }
55 void init() {
56     for(int i = 0; i < maxn; ++i) {
57         head[i] = -1;
58         dfn[i] = low[i] = 0;
59         uf[i] = i;
60     }
61     idx = tot = 0;
62     ans = INF;
63 }
64 int main() {
65     int u,v;
66     while(scanf("%d %d",&n,&m),n||m) {
67         init();
68         for(int i = 1; i <= m; ++i) {
69             scanf("%d %d %d",&u,&v,p+i);
70             add(u,v,i);
71             int tx = Find(u);
72             int ty = Find(v);
73             if(tx != ty) uf[tx] = ty;
74         }
75         int root = 0;
76         for(int i = 1; i <= n; ++i) if(uf[i] == i) root++;
77         if(root > 1){
78             printf("%d
",0);
79             continue;
80         }
81         for(int i = 1; i <= n; ++i)
82             if(!dfn[i]) tarjan(i,-1);
83         if(ans == INF) puts("-1");
84         else if(ans) printf("%d
",ans);
85         else puts("1");
86     }
87     return 0;
88 }

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