題意

輸入N，M，然後輸入M條邊（u，v，w）表示u-v有一條邊權為w的無向邊，求w最小的橋，輸出w。如果不存在橋輸出-1，圖不連通輸出0。

思路

模板題？就是有兩個坑點，①有重邊②圖不連通輸出0

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, num;
    Edge(int from, int to, int num): from(from), to(to), num(
 
num) {}
};
struct CUT
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[MAXN];
    int DFN[MAXN], LOW[MAXN], dfs_clock;   //時間戳，返祖邊到達的最小的祖先，計時
    int iscut[MAXN];                       //割點
    vector<Edge> cut;                      //橋

    void init(int n)
    {
        this->
 
n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }
    void AddEdge(int from, int to, int num)
    {
        edges.push_back(Edge(from, to, num));
        edges.push_back(Edge(to, from, num));
        m = edges.size();
        G[from].push_back(m-2);
        G[to]
 
.push_back(m-1);
    }
    int dfs(int u, int fa)
    {
        int lowu = DFN[u] = ++dfs_clock;
        int child = 0;
        bool flag = false;
        for (int i = 0; i < G[u].size(); i++)
        {
            Edge& e = edges[G[u][i]];
            int v = e.to;
            //重邊不算橋，只跳過第一條fa~u的邊；如果重邊也算橋就把所有的v==fa的邊跳過
            if (v == fa && !flag) { flag = true; continue; }
            if (!DFN[v])//樹邊
            {
                child++;
                int lowv = dfs(v, u);
                lowu = min(lowu, lowv);
                if (lowv >= DFN[u])//割點
                    iscut[u] = true;
                if (lowv > DFN[u])//橋
                    cut.push_back(e);
            }
            else //返祖邊
                lowu = min(lowu, DFN[v]);
        }
        if (fa < 0 && child == 1) iscut[u] = false;
        LOW[u] = lowu;
        return lowu;
    }
    int find_cut()
    {
        int cnt = 0;
        dfs_clock = 0;
        memset(DFN, 0, sizeof(DFN)), memset(LOW, 0, sizeof(LOW));
        cut.clear(), memset(iscut, 0, sizeof(iscut));
        for (int i = 0; i < n; i++)
            if (!DFN[i]) cnt++, dfs(i, -1);
        return cnt;
    }

}gao;

int n, m;
int main()
{
    while (~scanf("%d%d", &n, &m) && (n+m))
    {
        gao.init(n);
        while (m--)
        {
            int u, v, w; scanf("%d%d%d", &u, &v, &w);
            u--, v--;
            gao.AddEdge(u, v, w);
        }
        if (gao.find_cut() > 1) { printf("0\n"); continue; }
        int ans = INF;
        for (auto& i: gao.cut) ans = min(ans, i.num);
        if (ans == 0) ans = 1;
        if (ans == INF) ans = -1;
        printf("%d\n", ans);
    }   
    return 0;
}
/*
3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0
*/