hdu 4738 Caocao's Bridges(割边)

题目链接

用tarjan求桥上的最小权值

#include
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const int N=1e6+7;
const ll mod=1e9+7;
struct edge{
    int v,next,w;
};
edge e[N<<1];
int head[N],cnt = 1,dfn[N],low[N],num = 0,bridge[N];
void init(){
    memset(head,0,sizeof(head));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(bridge,0,sizeof(bridge));
    cnt=1; num=0;
}
void add(int u,int v,int w){
    e[++cnt].next=head[u]; e[cnt].v=v; e[cnt].w=w;
    head[u]=cnt;
}
void tarjan(int u,int edge){
    dfn[u]=low[u]=++num;
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].v; int w=e[i].w;
        if(!dfn[v]){
            tarjan(v,i);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
                bridge[i]=bridge[i^1]=1;
        }else if(i!=(edge^1))
            low[u]=min(low[u],low[v]);
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m; 
    while(cin>>n>>m){
        if(n==0&&m==0) break;
        init();
        for(int i=1;i<=m;i++){
            int u,v,w; cin>>u>>v>>w;
            add(u,v,w); add(v,u,w);
        }
        tarjan(1,-1);
        bool f=0;
        for(int i=1;i<=n;i++)
            if(!dfn[i]) f=1;
        if(f){
            cout<<"0\n";
            continue;
        }
        int ans=inf;
        for(int i=2;i<=cnt;i++){
            if(bridge[i])
                ans=min(ans,e[i].w);
        }
        if(ans==inf) cout<<"-1\n";
        else{
            if(ans==0) cout<<"1\n";
            else cout<"\n";
        } 
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wmj6/p/11153797.html