zoj2588 Burning Bridges(无向图的桥)

 

题目分析：就是求无向图的桥。

 

tarjan算法跑一遍，和无向图割点十分类似，这里要找low[v] > dfn[u]的边(u,v)便是割边，因为v是u的孩子，但是v无法访问到u的祖先，那么断开这条边原图必不连通，因此这是桥。这题会有平行边，平行边必定不是桥。所以dfs的时候要判断一下。

 

详情请见代码：

 

 

#include <iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;  
const int N = 10005;  
const int M = 500005;  
int m,n,num,ansnum,dfns;  
int head[N],ans[M],low[N],dfn[N];  
bool vis[N];  
struct node  
{  
    int to,next,id;  
}bridge[M<<1];  
void build(int s,int e,int id)  
{  
    bridge[num].id = id;  
    bridge[num].to = e;  
    bridge[num].next = head[s];  
    head[s] = num ++;  
}  
void dfs(int cur,int fa)  
{  
    vis[cur] = true;  
    int chongbian = 0;  
    dfn[cur] = low[cur] = dfns ++;  
    for(int i = head[cur];i != -1;i = bridge[i].next)  
    {  
        if(fa == bridge[i].to)  
            chongbian ++;  
        if(vis[bridge[i].to] == false)  
        {  
            dfs(bridge[i].to,cur);  
            low[cur] = min(low[cur],low[bridge[i].to]);  
            if(low[bridge[i].to] > dfn[cur])  
                ans[ansnum ++] = bridge[i].id;  
        }  
        else if(fa != bridge[i].to || chongbian > 1)  
                low[cur] = min(low[cur],dfn[bridge[i].to]);  
    }  
}  
void tarjan()  
{  
    int i;  
    dfns = 1;  
    memset(vis,false,sizeof(vis));  
    memset(dfn,0,sizeof(dfn));  
    for(i = 1;i <= n;i ++)  
        if(vis[i] == false)  
            dfs(i,-1);  
    printf("%d\n",ansnum);  
    sort(ans,ans + ansnum);  
    for(i = 0;i < ansnum;i ++)  
        printf(i == ansnum - 1?"%d\n":"%d ",ans[i]);  
}  
int main()  
{  
    int t,i;  
    int a,b;  
    scanf("%d",&t);  
    while(t --)  
    {  
        scanf("%d%d",&n,&m);  
        memset(head,-1,sizeof(head));  
        num = ansnum = 0;  
        for(i = 1;i <= m;i ++)  
        {  
            scanf("%d%d",&a,&b);  
            build(a,b,i);  
            build(b,a,i);  
        }  
        tarjan();  
        if(t)  
            puts("");  
    }  
    return 0;  
}  

 

补充：软件开发 , C++ ,