Break up

CF700C

首先考虑只能删一条边的做法，我们可以找出所有的桥，然后随便跑一条 S 到 T 路径，如果这条路径上有桥就说明可以，否则不行

发现这个做法其实是 O(M) 的

那么可以先随便找一条 N 到 M 的路径，分别尝试删这条路径上的边再套上面做法就好了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
//#define int long long
#define MAXN 3016 
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define inf 0x3f3f3f3f
#define cmx( a , b ) a = max( a , b )
#define cmn( a , b ) a = min( a , b )
int n , m , s , t;
int head[MAXN] , nex[MAXN * 50] , to[MAXN * 50] , wto[MAXN * 50] , ecn = -1;
void ade( int u , int v , int w ) {
    nex[++ecn] = head[u] , to[ecn] = v , wto[ecn] = w , head[u] = ecn;
}
int vis[MAXN] , pre[MAXN];
int fuck = 0x7f7f7f7f;
bool findpath( int u ) {
    if( u == t ) return true;
    vis[u] = 1;
    for( int i = head[u] ; ~i ; i = nex[i] ) {
        if( i == fuck || i == ( fuck ^ 1 ) ) continue;
        int v = to[i];
        if( vis[v] ) continue;
        pre[v] = i ^ 1;
        if( findpath( v ) ) return true;
    }
    return false;
}
int clo;
int dfn[MAXN] , low[MAXN];
int cut[MAXN * 50] , done[MAXN];
int use[MAXN * 50];
void tarjan( int u ) {
    done[u] = 1;
    dfn[u] = low[u] = ++ clo;
    for( int i = head[u] ; ~i ; i = nex[i] ) {
        if( use[i] ) continue;
        use[i] = use[i ^ 1] = 1;
        if( i == fuck || i == ( fuck ^ 1 ) ) continue;
        int v = to[i];
        if( !dfn[v] ) {
            tarjan( v );
            low[u] = min( low[u] , low[v] );
            if( low[v] > dfn[u] ) cut[i] = cut[i ^ 1] = true;
        } else if( done[v] == 1 )
            low[u] = min( low[u] , dfn[v] );
    }
    done[u] = 2;
}
vector<int> eds;
pii ans;
int main() {
    memset( head , -1 , sizeof head );
    cin >> n >> m >> s >> t;
    for( int i = 1 , u , v , w ; i <= m ; ++ i ) {
        scanf("%d%d%d",&u,&v,&w);
        ade( u , v , w ) , ade( v , u , w );
    }
    if( !findpath( s ) ) return puts("0") , puts("0") , 0;
    int c = t;
    while( c != s ) 
        eds.pb( pre[c] ) , c = to[pre[c]];
    int res = 0x7f7f7f7f;
    for( int i = 0 ; i < eds.size() ; ++ i ) {
        fuck = eds[i];
        memset( cut , 0 , sizeof cut );
        memset( done , 0 , sizeof done );
        memset( use , 0 , sizeof use );
        memset( dfn , 0 , sizeof dfn );
        memset( low , 0 , sizeof low );
        clo = 0;
        tarjan( s );
        memset( vis , 0 , sizeof vis );
        if( ! findpath( s ) ) {
            if( res > wto[fuck] ) 
                res = wto[fuck] ,
                ans = mp( 0 , fuck >> 1 );
            continue;
        }
        c = t;
        while( c != s ) { 
            if( cut[pre[c]] ) 
                if( res > wto[fuck] + wto[pre[c]] )
                    res = wto[fuck] + wto[pre[c]] , ans = mp( fuck >> 1 , pre[c] >> 1 );
            c = to[pre[c]];
        }
    }
    if( res == 0x7f7f7f7f ) return puts("-1") , 0;
    if( ! ans.fi ) {
        printf("%d\n",wto[ans.se << 1]);
        puts("1");
        printf("%d",ans.se + 1);
    } else {
        printf("%d\n",res);
        puts("2");
        printf("%d %d" , ans.fi + 1 , ans.se + 1);
    }
}