Break up CF700C

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);
	}
}
posted @ 2019-10-11 08:16  yijan  阅读(268)  评论(0编辑  收藏  举报