HDU 4738 Caocao's Bridges taijan （求割边，神坑）

神坑题。这题的坑点有1.判断连通，2.有重边，3.至少要有一个人背炸药

因为有重边，tarjan的时候不能用子结点和父节点来判断是不是树边的二次访问，所以我的采用用前向星存边编号的奇偶性关系，用^1来判断是不是树边

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxe = 1000005<<1;
const int maxv = 1005;
int clock;
int dfn[maxv],low[maxv],ecnt;
int head[maxv],nxt[maxe],to[maxe],wei[maxe];

void init()
{
    clock = ecnt = 0;
    memset(head,-1,sizeof(head));
    memset(dfn,0,sizeof(dfn));
}

void addEdge(int u,int v,int w)
{
    to[ecnt] = v;
    wei[ecnt] = w;
    nxt[ecnt] = head[u];
    head[u] = ecnt++;
}

int ans;

void tarjan(int u,int fa)
{
    dfn[u] = low[u] = ++clock;
    for(int i = head[u]; ~i ; i = nxt[i]){
        int v = to[i];
        if(i == (fa^1)) continue;
        if(!dfn[v]){
            tarjan(v,i);
            low[u] = min(low[u],low[v]);
            if(low[v] > dfn[u]){
                ans = min(ans,wei[i]);
            }
        }else  {
            low[u] = min(low[u],dfn[v]);
        }
    }
}


int sum;
bool vis[maxv];
void dfs(int u){
    if(vis[u]) return;
    sum++;
    vis[u] = 1;
    for(int i = head[u]; ~i ; i = nxt[i]){
        dfs(to[i]);
    }
}
const int INF = 1e9;
int main()
{
    int N,M;
    while(~scanf("%d%d",&N,&M)&&N){
        init();
        for(int i = 0; i < M; i++){
            int U,V,W;
            scanf("%d%d%d",&U,&V,&W);
            addEdge(U,V,W);
            addEdge(V,U,W);
        }
        memset(vis,0,sizeof(vis));
        sum = 0;
        dfs(1);
        if(sum < N) { printf("0\n"); continue; }
        ans = INF;
        tarjan(1,-1);
        if(ans == INF) { printf("-1\n"); continue; }
        ans = max(ans,1);
        printf("%d\n",ans);
    }
    return 0;
}