hdu 4738（双联通缩点）

来源：互联网发布：日本知乎编辑：程序博客网时间：2024/06/05 06:19

此题有坑！！！

桥的边权有可能是0，但是这时候要输出1，因为周瑜至少的排一个人去炸呀，那个年代又没有导弹！！！卧槽，有时候还得联系实际！

#include<cstdio>#include<algorithm>#include<cstring>#include<stack>#include<vector>using namespace std;const int inf = 1<<20;const int maxn = 1005;int n,m;struct node{    int to;    int next;    int w;    node(){}    node(int a,int b,int c):to(a),next(b),w(c){}}edge[1002*1002];struct Edge{    int to;    int w;    Edge(int a,int b):to(a),w(b){}};vector<Edge>G[maxn];int head[maxn],dfn[maxn],low[maxn],belong[maxn];int tot,ti,cnt;stack<int>s;void tarjan(int u,int fa){    dfn[u] = ++ti; low[u] = ti;    s.push(u); int flag = 0;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v = edge[i].to;        if(v==fa&&flag==0){flag=1;continue;}        if(!dfn[v])        {            tarjan(v,u);            low[u] = min(low[u],low[v]);        }        else if(belong[v]==-1)        {            low[u] = min(dfn[v],low[u]);        }    }    if(dfn[u]==low[u])    {        int y; cnt++;        do{            y = s.top(); s.pop();            belong[y] = cnt;        }while(y!=u);    }}int res;void dfs(int u,int fa){    int len = G[u].size();    for(int i=0;i<len;i++)    {        int v = G[u][i].to;        if(v==fa) continue;        res = min(res,G[u][i].w);        dfs(v,u);    }}void cbh(int u){    dfn[u] = 1;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v = edge[i].to;        if(!dfn[v])        {            dfn[v] = 1;            cbh(v);        }    }}int main(){    int u,v,w;    while(scanf("%d%d",&n,&m))    {        if(n==0&&m==0) break;        tot = 0;        memset(head,-1,sizeof(head));        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&w);            if(u==v) continue;            edge[tot] = node(v,head[u],w);            head[u] = tot++;            edge[tot] = node(u,head[v],w);            head[v] = tot++;        }        int key = 0;        memset(dfn,0,sizeof(dfn));        for(int i=1;i<=n;i++)        {            if(!dfn[i]){key++;cbh(i);}        }        if(key>1){printf("0\n");continue;}        memset(dfn,0,sizeof(dfn));        memset(low,0,sizeof(low));        memset(belong,-1,sizeof(belong));        ti = 0; cnt = 0;        for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i,-1);        if(cnt==1) {printf("-1\n"); continue; }        for(int i=1;i<=n;i++) G[i].clear();        for(int i=1;i<=n;i++)        {            for(int j=head[i];j!=-1;j=edge[j].next)            {                if(belong[i]!=belong[edge[j].to])                {                    G[belong[i]].push_back(Edge(belong[edge[j].to],edge[j].w));                }            }        }        res = inf;        dfs(1,-1);        if(res==0) res++;        printf("%d\n",res);    }    return 0;}

阅读全文

0 0