zoj2588—Burning Bridges （割边的求解）

题目链接：传送门

Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very beautiful and are loved by everyone in the kingdom. Of course, the system of bridges is designed in such a way that one can get from any island to any other one.

But recently the great sorrow has come to the kingdom. Ferry Kingdom was conquered by the armies of the great warrior Jordan and he has decided to burn all the bridges that connected the islands. This was a very cruel decision, but the wizards of Jordan have advised him no to do so, because after that his own armies would not be able to get from one island to another. So Jordan decided to burn as many bridges as possible so that is was still possible for his armies to get from any island to any other one.

Now the poor people of Ferry Kingdom wonder what bridges will be burned. Of course, they cannot learn that, because the list of bridges to be burned is kept in great secret. However, one old man said that you can help them to find the set of bridges that certainly will not be burned.

So they came to you and asked for help. Can you do that?

Input

The input contains multiple test cases. The first line of the input is a single integer T (1 <= T <= 20) which is the number of test cases. T test cases follow, each preceded by a single blank line.

The first line of each case contains N and M - the number of islands and bridges in Ferry Kingdom respectively (2 <= N <= 10 000, 1 <= M <= 100 000). Next M lines contain two different integer numbers each and describe bridges. Note that there can be several bridges between a pair of islands.

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Output

On the first line of each case print K - the number of bridges that will certainly not be burned. On the second line print K integers - the numbers of these bridges. Bridges are numbered starting from one, as they are given in the input.

Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.

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Sample Input

26 71 22 32 45 41 34 53 610 162 63 76 55 95 41 29 86 42 103 87 91 42 410 51 66 10

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Sample Output

23 714 

解题思路：和割点的解法差不多，只是割边的判断条件改为—当(u,v)为生成树的边，且满足dfn[u]<low[v]时(u,v)时割边

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <stack>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;typedef pair<int,int>PA;const int N = 100900;const int M = 10900;const int INF = 0x3fffffff;const double eps = 1e-8;const double PI = acos(-1.0);//dfn[u]是深度优先搜索树节点u的优先级//low[u]是从u或者u的子孙出发通过回边可以到达的最低深度优先数//low[u] = min{ dfn[u] ,//              min{ low[w] | w是u的一个子女 } ,//              min{ dfn[v] | v与u邻接，且(u,v)是一条回边} }/*    割边的判断:    无向图中的一条边(u,v)是割边，当且仅当(u,v)是生成树种的边，且满足dfn[u]<low[v]    割边的求解与割点的求解类似*///向下搜索时，如果顶点v是u的相邻顶点，若v还未被访问，则v是u的儿子节点//若v被访问了，则v是u的祖先节点，且(u,v)是一条回边struct Edge{    int node;    Edge*next;}m_edge[N*2];Edge*head[M];int low[M],dfn[M],Flag[M],Ecnt,cnt;vector<PA>st;void init(){    Ecnt = cnt = 0;    fill( Flag , Flag+M , 0 );    fill( head , head+M , (Edge*)0 );}void mkEdge( int a , int b ){    m_edge[Ecnt].node = b;    m_edge[Ecnt].next = head[a];    head[a] = m_edge+Ecnt++;}void tarjan( int u , int father ){    Flag[u] = 1;    low[u] = dfn[u] = cnt++;    for( Edge*p = head[u] ; p ; p = p->next ){        int v = p->node;        if( !Flag[v] ){            tarjan(v,u);            low[u] = min(low[u],low[v]);            //判断(u,v)是否是割边            if( low[v] > dfn[u] ){                st.push_back(PA(u,v));            }        }        //v是u的祖先，且(u,v)是一条回边        if( v != father && Flag[v] )            low[u] = min(low[u],dfn[v]);    }}map<PA,int>mp;void Build( int m ){    int a,b,num = 0;    for( int i = 0 ; i < m ; ++i ){        scanf("%d%d",&a,&b);        if( mp[PA(a,b)] != 0 ){            //将重边标记为无穷大，重边不可能是割边            mp[PA(a,b)] = INF;            mp[PA(b,a)] = INF;            num += 2;        }else{            mp[PA(a,b)] = (++num);            mp[PA(b,a)] = (++num);            mkEdge(a,b);            mkEdge(b,a);        }    }}int main(){    int T,cas = 0;    scanf("%d",&T);    while( T-- ){        init();        if( !mp.empty() ) mp.clear();        while( !st.empty() ) st.pop_back();        int n,m;        scanf("%d%d",&n,&m);        Build(m);        tarjan(1,-1);        int rec[N],Count = 0;        for( int i = 0 ; i < st.size() ; ++i ){            int s = mp[st[i]];            //重边不可能是割边            if( s != INF ) rec[Count++] = (s+1)/2;        }        if( cas != 0 ) printf("\n");        printf("%d\n",Count);        if( Count > 0 ){            sort( rec , rec+Count );            printf("%d",rec[0]);            for( int i = 1 ; i < Count ; ++i ){                printf(" %d",rec[i]);            }            printf("\n");        }        ++cas;    }    return 0;}