Critical Links-UVa796(无向图求桥)

 

 

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=737

 

 

桥:
一个连通图,无向连通图中,如果删除某条边后,图变成不连通了,则该边为桥;
求桥:
在求割点的基础上吗,假如一个边没有重边(重边 1-2, 1->2 有两次,那么 1->2 就是有两条边了,那么 1->2就不算是桥了)。
当且仅当 (u,v) 为父子边,且满足 dfn[u] < low[v]
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>

using namespace std;
#define N 200

int low[N],dfn[N],n,fa[N],Stack[N];
int Time,top,ans[N];
struct node
{
    int x,y;
}a[N];
vector<vector <int> >G;

void Inn()
{
    G.clear();
    G.resize(n+1);
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(fa,0,sizeof(fa));
    memset(ans,0,sizeof(ans));
    memset(Stack,0,sizeof(Stack));
    Time=top=0;
}
bool cmp(node c,node d)
{
    if(c.x!=d.x)
        return c.x<d.x;
    else
        return c.y<d.y;
}
void Tarjin(int u,int f)
{
    low[u]=dfn[u]=++Time;
    Stack[top++]=u;
    fa[u]=f;
    int len, v;
    len=G[u].size();
    for(int i=0; i<len; i++)
    {
        v=G[u][i];
        if(!dfn[v])
        {
            Tarjin(v,u);
            low[u]=min(low[u],low[v]);
        }
        else if(f!=v)
            low[u]=min(low[u],dfn[v]);
    }
}

void slove()
{
    int ans=0,i;
    for(i=0; i<n; i++)
    {
        if(!dfn[i])
            Tarjin(i,-1);
    }
    for(i=0; i<n; i++)
    {
        int v=fa[i];
        if(v!=-1 && low[i]>dfn[v])
        {
            a[ans].x=i;
            a[ans].y=v;
            int t;
            if(a[ans].x>a[ans].y)
            {
                t=a[ans].x;
                a[ans].x=a[ans].y;
                a[ans].y=t;
            }
            ans++;
        }

    }
    sort(a,a+ans,cmp);
    printf("%d critical links\n",ans);
    for(i=0; i<ans; i++)
        printf("%d - %d\n",a[i].x,a[i].y);
    printf("\n");
}

int main()
{
    int h,m,b,i;
    while(scanf("%d",&n)!=EOF)
    {
        Inn();
        for(i=0; i<n; i++)
        {
            scanf("%d (%d)",&h,&m);
            while(m--)
            {
                scanf("%d",&b);
                G[h].push_back(b);
                G[b].push_back(h);
            }
        }
        slove();
    }
    return 0;
}

/*
8
0 (1) 1
1 (3) 2 0 3
2 (2) 1 3
3 (3) 1 2 4
4 (1) 3
7 (1) 6
6 (1) 7
5 (0)
*/

 

posted @ 2015-10-28 16:23  啦咯  阅读(303)  评论(0编辑  收藏  举报