【代码】POJ 3352

// 题目来源：POJ 3352 ( CCC 2007 )// 题目大意：给定无向图G，为了满足删去任意一条边后原图任然连通，则原图至少需增加多少条边// 解决方法：求出原图的桥数量，记作sum，则（sum+1）/2即为答案#include <cstdio>void link( int, int );void tarjan( int, int );void dfs( int );int next[2002], h[2002], g[2002], code[2002], id[2002], l[2002], r[2002], dfn[2002], low[2002];int n, m, t = 1, cnt, bcnt, sum, index;bool p[2002];int main( ){    freopen( "input.txt", "r", stdin );    freopen( "output.txt", "w", stdout );    int aa, bb;    scanf( "%d%d", &n, &m );    while( m-- )    {        scanf( "%d%d", &aa, &bb );        link( aa, bb );    }    tarjan( 1, 0 );    for( int i = 1; i <= n; i++ )        if( code[i] == 0 )        {            cnt++;            dfs( i );        }    for( int i = 1; i <= bcnt; i++ )    {        id[code[l[i]]]++;        id[code[r[i]]]++;    }    for( int i = 1; i <= cnt; i++ )        if( id[i] == 1 ) sum++;    printf( "%d", (sum+1)/2 );    return 0;}void link( int aa, int bb ){    next[++t] = h[aa];    h[aa] = t;    g[t] = bb;    next[++t] = h[bb];    h[bb] = t;    g[t] = aa;}void tarjan( int i, int num ){    int j;    dfn[i] = low[i] = ++index;    for( int k = h[i]; k; k = next[k] )    {        j = g[k];        if( ( k^num ) == 1 ) continue;        if( !dfn[j] )        {            tarjan( j, k );            if( low[j] < low[i] ) low[i] = low[j];            if( dfn[i] < low[j] )            {                 p[k] = p[k^1] = 1;                l[++bcnt] = i;                r[bcnt] = j;            }        }        else            if( dfn[j] < low[i] ) low[i] = dfn[j];    }}void dfs( int i ){    int j;    code[i] = cnt;    for( int k = h[i]; k; k = next[k] )    {        j = g[k];        if( p[k] || code[j] != 0 ) continue;        dfs( j );    }}