hdu-4738.Caocao's Bridges(鍥句腑鏉冨€兼渶灏忕殑妗?

Posted 2020-11-12

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Caocao鈥榮 Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10933    Accepted Submission(s): 3065

Problem Description

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn鈥榯 give up. Caocao鈥榮 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鈥榮 army could easily attack Zhou Yu鈥榮 troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao鈥榮 army could be deployed very conveniently among those islands. Zhou Yu couldn鈥榯 stand with that, so he wanted to destroy some Caocao鈥榮 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鈥榯 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鈥榯 succeed any way, print -1 instead.

 

 

Sample Input

3 3 1 2 7 2 3 4 3 1 4 3 2 1 2 7 2 3 4 0 0

 

 

Sample Output

-1 4

 

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

 

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  1 /*************************************************************************
  2     > File Name: hdu-4738.caocaos_bridges.cpp
  3     > Author: CruelKing
  4     > Mail: [e-mail] 
  5     > Created Time: 2019骞?9鏈?7鏃?鏄熸湡鍏?21鏃?1鍒?1绉?
  6     鏈鎬濊矾:鏃犲悜鍥炬墍鏈夋ˉ涓潈鍊肩殑閭ｆ潯妗ョ殑鏉冨€?
  7     娉ㄦ剰:鏈夐噸杈?濡傛灉妗ヤ笂娌℃晫浜?闇€瑕佹湁浜烘姉tnt,鍥犳闇€瑕佽緭鍑?.
  8     濡傛灉鍒濆鍥句笉杩為€氬垯杈撳嚭0.
  9  ************************************************************************/
 10 
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <map>
 14 using namespace std;
 15 
 16 const int maxn = 1000 + 5, maxm = maxn * maxn + 5, inf = 0x3f3f3f3f;
 17 int n, m;
 18 int tot, head[maxn];
 19 
 20 int bridge, top, Index, min_bridge;
 21 int dfn[maxn], low[maxn], stack[maxn];
 22 bool instack[maxn];
 23 
 24 map<int, int> mp;
 25 
 26 struct Edge {
 27     int to, cost, next;
 28     bool cut;
 29 } edge[maxm << 1];
 30 
 31 int min(int x, int y) {
 32     return x > y ? y : x;
 33 }
 34 
 35 void init() {
 36     mp.clear();
 37     memset(head, -1, sizeof head);
 38     tot = 0;
 39 }
 40 
 41 void addedge(int u, int v ,int w) {
 42     edge[tot] = (Edge){v, w, head[u], false}; head[u] = tot ++;
 43     edge[tot] = (Edge){u, w, head[v], false}; head[v] = tot ++;
 44 }
 45 
 46 bool ishash(int u, int v) {
 47     return mp[u * maxn + v] ++ || mp[v * maxn + u] ++;
 48 }
 49 
 50 void tarjan(int u, int pre) {
 51     int v;
 52     stack[top ++] = u;
 53     instack[u] = true;
 54     dfn[u] = low[u] = ++ Index;
 55     int pre_cnt = 0;
 56     for(int i = head[u]; ~i; i = edge[i].next) {
 57         v = edge[i].to;
 58         if(v == pre && pre_cnt == 0) {
 59             pre_cnt ++;
 60             continue;
 61         }
 62         if(!dfn[v]) {
 63             tarjan(v, u);
 64             if(low[u] > low[v]) low[u] = low[v];
 65             if(low[v] > dfn[u]) {
 66                 edge[i].cut = true;
 67                 edge[i ^ 1].cut = true;
 68                 min_bridge = min(min_bridge, edge[i].cost);
 69                 bridge ++;
 70             }
 71         } else if(low[u] > dfn[v]) low[u] = dfn[v];
 72     }
 73     top --;
 74     instack[u] = false;
 75 }
 76 
 77 void solve() {
 78     memset(instack, false, sizeof instack);
 79     memset(dfn, 0, sizeof dfn);
 80     memset(low, 0, sizeof low);
 81     top = Index = bridge = 0;
 82     min_bridge = inf;
 83     for(int i = 1; i <= n; i ++) {
 84         if(!dfn[i]) {
 85             tarjan(i, i);//cnt ++;
 86         }
 87     }
 88     if(min_bridge == inf) min_bridge = -1;
 89     else if(min_bridge == 0) min_bridge = 1;//if cnt != 1 : min_bridge = 0;
 90     printf("%d
", min_bridge);
 91 }
 92 
 93 int fa[maxn];
 94 
 95 int find(int x) {
 96     if(fa[x] != x) return fa[x] = find(fa[x]);
 97     else return x;
 98 }
 99 
100 void unionset(int u, int v) {
101     u = find(u);
102     v = find(v);
103     if(u != v) fa[u] = v;
104 }
105 
106 int main() {
107     int u, v, w;
108     while(~scanf("%d %d", &n, &m) && (n || m)) {
109         init();
110         for(int i = 1; i <= n; i ++) fa[i] = i;
111         for(int i = 0; i < m; i ++) {
112             scanf("%d %d %d", &u, &v, &w);
113 //            if(ishash(u, v)) continue;
114             addedge(u, v, w);
115             unionset(u, v);
116         }
117         bool flag = true;
118         for(int i = 1; i <= n; i ++)
119             if(find(i) != find(1)) {
120                 flag = false;
121                 break;
122             }
123         if(flag)
124             solve();
125         else printf("0
");
126     }
127     return 0;
128 }