HDU Caocao's Bridges

题意挺简单的但是有好多细节要处理好，就是让求所有割边最小权值
1）自环
2）如果地图根本不是一个联通图那么我们根本不需要炸，答案就是0
3）如果路上的侍卫是0，那么我们至少需要一个人去被炸， 此时答案是1
4）其他情况就正常来就行，求割边

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>using namespace std;#define MAXN 3005#define MAXE 3000005#define INF 1000000007struct Edge{    int u, v, next, val;}e[MAXE];int head[MAXE], cnt, ans;int dfn[MAXN], low[MAXN], fa[MAXN], dep;bool bri[MAXE];void init(){    memset(head, -1, sizeof(head));    memset(fa, -1, sizeof(fa));    memset(dfn, 0, sizeof(dfn));    memset(bri, 0, sizeof(bri));    dep = 0, cnt = 0;    ans = INF;}void Addedge(int uu, int vv, int w){    e[cnt].u = uu, e[cnt].v = vv, e[cnt].val = w;    e[cnt].next = head[uu], head[uu] = cnt++;}int findfa(int x){    if(fa[x] == -1) return x;    return fa[x] = findfa(fa[x]);}void unit(int x, int y){    int fax = findfa(x);    int fay = findfa(y);    if(fax != fay)         fa[fax] = fay;}void dfs( int u, int pre){    dfn[u] = low[u] = ++dep;    //cout<<"u"<<" "<<u<<endl;    for(int i = head[u]; i != -1; i = e[i].next)    {        int v = e[i].v;        if(i == (pre^1)) continue;        if(!dfn[v])        {            dfs(v, i);            low[u] = min(low[u], low[v]);            if(low[v] > dfn[u])                ans = min(ans, e[i].val);        }        else            low[u] = min(low[u], dfn[v]);    }}void solve( int n, int m){    dfs(1, -1);    if(ans == INF)        printf("-1\n");    else        printf("%d\n",ans);}int main(){    int n, m, val, u, v;    while(scanf("%d %d",&n, &m) != EOF &&(n + m))    {        init();        for( int i = 0; i < m; i++)        {            scanf("%d %d %d",&u, &v, &val);            if( u == v) continue;            if(!val) val = 1;            Addedge(u, v, val);            Addedge(v, u, val);            unit(u, v);        }        int i;        for( i = 2; i <= n; i++)        {            if(findfa(i) != findfa(1))                 break;        }        if(i <= n)          printf("0\n");        else         solve(n, m);    }}