HDU-4738-Caocao's Bridges（割边）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4738

大致题意：无向边，带权值，让求权值最小的一个割边

注意以下几点：

1，多图：ans=0

2，割边最小权值=0 ：ans=1（至少派一个士兵去炸桥）

3，无割边： ans=-1

4，重边的处理

//#include <bits/stdc++.h>#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int Scan(){    int res=0,ch,flag=0;    if((ch=getchar())=='-')flag=1;    else if(ch>='0'&&ch<='9')res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';    return flag?-res:res;}void Out(int a){    if(a>9)Out(a/10);    putchar(a%10+'0');}#define INF 0x3f3f3f3f#define bug cout<<"bug"<<endlconst int MAXN = 1e5+7;struct node{    int v,next,cost;    int vis;} edge[MAXN<<3];int head[MAXN],index;void add_edge(int u, int v, int cost){    edge[index].vis=0;    edge[index].v=v;    edge[index].cost=cost;    edge[index].next=head[u];    head[u]=index++;    edge[index].vis=0;    edge[index].v=u;    edge[index].cost=cost;    edge[index].next=head[v];    head[v]=index++;}int belong[MAXN],DFN[MAXN],low[MAXN],stack_[MAXN];int cir,deep,top,temp;bool in_stack[MAXN],use[MAXN];void Tarjan(int u){    int p;    DFN[u]=low[u]=++deep;    in_stack[u]=1;    stack_[top++]=u;    temp++;    for(int i=head[u]; i+1; i=edge[i].next)    {        int v=edge[i].v;        if(edge[i].vis)continue;        edge[i].vis=edge[i^1].vis=1;        if(!DFN[v])        {            Tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(in_stack[v])            low[u]=min(low[u],DFN[v]);    }    if(DFN[u]==low[u])    {        cir++;        do{            p=stack_[--top];            in_stack[p]=0;            belong[p]=cir;        }        while(p!=u);    }}int main(){    int n,m,k,a,b,c;    int cas=1;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(!n)break;        temp=cir=index=deep=top=0;        memset(DFN,0,sizeof(DFN));        memset(head,-1,sizeof(head));        for(int i=0; i<m; ++i)        {            scanf("%d%d%d",&a,&b,&c);            add_edge(a,b,c);        }        Tarjan(1);        if(temp<n){printf("0\n");continue;}        int ans=INF,flag=1;        for(int i=1; i<=n; ++i)            for(int j=head[i]; j+1; j=edge[j].next)            {                int v=edge[j].v;                if(belong[i]!=belong[v] && edge[j].cost<ans)                {                    flag=0;                    ans=edge[j].cost;                }            }        if(ans==0)ans=1;        if(flag)printf("-1\n");        else printf("%d\n",ans);    }    return 0;}/*4 31 2 11 2 33 4 44 41 2 11 2 33 4 42 4 6*/

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