CF - 652 E Pursuit For Artifacts 边双联通

题目传送门

题解总结起来其实很简单。

把所有的边双联通分量缩成一个点，然后建立好新边， 然后再从起点搜到终点就好了。

 

代码：

/*
code by: zstu wxk
time: 2019/02/23
*/
#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 3e5 + 100;
const int M = N * 2;
int n, m;
int head[N], to[M], nt[M], val[M], bridge[M];
int tot;
void add(int u, int v, int w){
    to[tot] = v;
    nt[tot] = head[u];
    val[tot] = w;
    head[u] = tot++;
}
int dfn[N], low[N], dtot;
void Tarjan(int o, int u){
    dfn[u]= low[u] = ++dtot;
    for(int i = head[u]; ~i; i = nt[i]){
        int v = to[i];
        if(!dfn[v]){
            Tarjan(u, v);
            low[u] = min(low[u], low[v]);
            if(low[v] > dfn[u])
                bridge[i] = bridge[i^1] = 1;
        }
        else if(v != o)
            low[u] = min(low[u], dfn[v]);
    }
}
int c[N], dcc;
void dfs(int u){
    c[u] = dcc;
    for(int i = head[u]; i; i = nt[i]){
        int v = to[i];
        if(c[v] || bridge[i]) continue;
        dfs(v);
    }
}
int ok[N];
vector<pll> vc[N];
void e_dcc(){
    for(int i = 1; i <= n; ++i)
        if(!dfn[i]) Tarjan(0, i);
    for(int i = 1; i <= n; ++i)
        if(!c[i]) {
            ++dcc;
            dfs(i);
        }
    for(int i = 0; i <= tot; i += 2){
        int u = to[i^1], v = to[i];
        u = c[u], v = c[v];
        if(u == v){
            ok[u] |= val[i];
        }
        else {
            vc[u].pb({v,val[i]});
            vc[v].pb({u,val[i]});
        }
    }
}
int flag = 0;
int t;
void Dfs(int o, int u, int f){
    if(u == t){
        flag |= f;
        return ;
    }
    for(pll x : vc[u]){
        int v = x.fi;
        if(v == o) continue;
        Dfs(u, v, f | x.se | ok[v]);
    }
}
void Ac(){
    int u, v, w;
    memset(head, -1, sizeof head);
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w); add(v, u, w);
    }
    e_dcc();
    scanf("%d%d", &u, &v);
    u = c[u];
    t = c[v];
    Dfs(0, u, ok[u]);
    if(flag) puts("YES");
    else puts("NO");
}
int main(){
    while(~scanf("%d%d", &n, &m)){
        Ac();
    }
    return 0;
}