UVA 796 Critical Links(无向图求桥)

题目大意：给你一个网络要求这里面的桥。

输入数据：

n 个点

点的编号  （与这个点相连的点的个数m）  依次是m个点的

 

输入到文件结束。

桥输出的时候需要排序

 

知识汇总：

桥：   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]

这里对重边处理的时候用了两种方法。写了两个代码，也挺简单的。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <cstring>
using namespace std;
#define INF 0xfffffff
#define maxn 11005
#define min(a,b) (a<b?a:b)
struct node
{
    int x, y;
    bool friend operator < (node A,node B)
    {
        if(A.x == B.x)
            return A.y < B.y;
        return A.x < B.x;
    }
}bridge[maxn];
int n, dfn[maxn], low[maxn], Father[maxn], Time;
vector<int> G[maxn];

void init()
{
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(Father, 0, sizeof(Father));
    Time = 0;
    for(int i=0; i<n; i++)
        G[i].clear();
}

void Tarjan(int u,int fa)
{
    Father[u] = fa;
    low[u] = dfn[u] = ++Time;
    int len = G[u].size(), v, k = 0;

    for(int i=0; i<len; i++)
    {
        v = G[u][i];

        if(v == fa && !k)
        {
            k ++;
            continue;
        }
        if( !low[v] )
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}

void solve()
{
    int ans = 0;
    for(int i=0; i<n; i++)
    {
        if(!dfn[i])
            Tarjan(i,-1);
    }


    for(int i=0; i<n; i++)
    {
        int v = Father[i];
        if(dfn[v] < low[i] && v != -1)
        {

            bridge[ans].x = i;
            bridge[ans].y = v;

            if(bridge[ans].x > bridge[ans].y)
                swap(bridge[ans].x, bridge[ans].y);
            ans ++;
        }
    }
    sort(bridge, bridge + ans);

    printf("%d critical links
", ans);

    for(int i=0; i<ans; i++)
    {
        printf("%d - %d
",bridge[i].x,bridge[i].y);
    }
    printf("
");
}

int main()
{
    while(scanf("%d",&n) != EOF)
    {
        init();
        for(int i=0; i<n; i++)
        {
            int a, b, m;
            scanf("%d (%d)",&a,&m);

            while(m--)
            {
                scanf("%d", &b);
                G[a].push_back(b);
               // G[b].push_back(a);
            }
        }
        solve();
    }
    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)
*/

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <cstring>
usingnamespace std;
#define INF 0xfffffff
#define maxn 11005
#define min(a,b) (a<b?a:b)
/** 无向图求桥 **/struct node
{
    int x, y;
    bool friend operator < (node A,node B)
    {
        if(A.x == B.x)
            return A.y < B.y;
        return A.x < B.x;
    }
}bridge[maxn];
int n, dfn[maxn], low[maxn], Father[maxn], Time;
vector<int> G[maxn];

void init()
{
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(Father, 0, sizeof(Father));
    Time = 0;
    for(int i=0; i<n; i++)
        G[i].clear();
}

void Tarjan(int u,int fa)
{
    Father[u] = fa;
    low[u] = dfn[u] = ++Time;
    int len = G[u].size(), v;

    for(int i=0; i<len; i++)
    {
        v = G[u][i];

        if( !low[v] )
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        elseif(fa != v)
            low[u] = min(low[u], dfn[v]);
    }
}

void solve()
{
    int ans = 0;
    for(int i=0; i<n; i++)
    {
        if(!low[i])
            Tarjan(i, -1);
    }

    for(int i=0; i<n; i++)
    {
        int v = Father[i];
        if(v != -1 && dfn[v] < low[i])
        {

            bridge[ans].x = i;
            bridge[ans].y = v;

            if(bridge[ans].x > bridge[ans].y)
                swap(bridge[ans].x, bridge[ans].y);
            ans ++;
        }
    }
    sort(bridge, bridge + ans);

    printf("%d critical links
", ans);

    for(int i=0; i<ans; i++)
    {
        printf("%d - %d
",bridge[i].x,bridge[i].y);
    }
    printf("
");
}

int main()
{
    while(scanf("%d",&n) != EOF)
    {
        init();
        for(int i=0; i<n; i++)
        {
            int a, b, m;
            scanf("%d (%d)",&a,&m);

            while(m--)
            {
                scanf("%d", &b);
                G[a].push_back(b);
                G[b].push_back(a);
            }
        }
        solve();
    }
    return0;
}