Educational Codeforces Round 10 E.Pursuit For Artifacts

  
  
  
  

   
   
   
   
 
    
 
     
 
      
 
       
 
        
 
         E. Pursuit For Artifacts 
        
 
        
 
          
         

           time limit per test 
         
3 seconds 
        
 
        
 
          
         

           memory limit per test 
         
512 megabytes 
        
 
        
 
          
         

           input 
         
standard input 
        
 
        
 
          
         

           output 
         
standard output 
        
 
       
 
       
 
        
Johnny is playing a well-known computer game. The game are in some country, where the player can freely travel, pass quests and gain an experience.
 
        
In that country there are n islands and m bridges between them, so you can travel from any island to any other. In the middle of some bridges are lying ancient powerful artifacts. Johnny is not interested in artifacts, but he can get some money by selling some artifact.
 
        
At the start Johnny is in the island a and the artifact-dealer is in the island b (possibly they are on the same island). Johnny wants to find some artifact, come to the dealer and sell it. The only difficulty is that bridges are too old and destroying right after passing over them. Johnnie's character can't swim, fly and teleport, so the problem became too difficult.
 
        
Note that Johnny can't pass the half of the bridge, collect the artifact and return to the same island.
 
        
Determine if Johnny can find some artifact and sell it.
 
       
 
       
 
        
 
         Input 
        
 
        
The first line contains two integers n and m (1 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of islands and bridges in the game.
 
        
Each of the next m lines contains the description of the bridge — three integers xi, yi, zi (1 ≤ xi, yi ≤ n, xi ≠ yi, 0 ≤ zi ≤ 1), where xi and yiare the islands connected by the i-th bridge, zi equals to one if that bridge contains an artifact and to zero otherwise. There are no more than one bridge between any pair of islands. It is guaranteed that it's possible to travel between any pair of islands.
 
        
The last line contains two integers a and b (1 ≤ a, b ≤ n) — the islands where are Johnny and the artifact-dealer respectively.
 
       
 
       
 
        
 
         Output 
        
 
        
If Johnny can find some artifact and sell it print the only word "YES" (without quotes). Otherwise print the word "NO" (without quotes).
 
       
 
       
 
        
 
         Examples 
        
 
        
 
         
 
          
 
           input 
          
 
          
6 7
1 2 0
2 3 0
3 1 0
3 4 1
4 5 0
5 6 0
6 4 0
1 6
 
         
 
         
 
          
 
           output 
          
 
          
YES
 
         
 
         
 
          
 
           input 
          
 
          
5 4
1 2 0
2 3 0
3 4 0
2 5 1
1 4
 
         
 
         
 
          
 
           output 
          
 
          
NO
 
         
 
         
 
          
 
           input 
          
 
          
5 6
1 2 0
2 3 0
3 1 0
3 4 0
4 5 1
5 3 0
1 2
 
         
 
         
 
          
 
           output 
          
 
          
YES
 
          
 
           
 
          
 
         
 
        
 
       
 
      
 
     
 
    
 
   
  
  
  
  

题意:有n个点,m条边,接下来有m条边,如果他是特殊边边长就为1,否则为0,求从ST到ED每条边最多经过一次能否经过任一特殊边。

 

思路:边双连通缩点,如果缩点之后的环中有特殊边,那么把这个点的值标记为1,否则为0,因为缩点之后会形成一棵树,所以任意两点之间如果有路径,那么一定是唯一的,而且这题的路径有要求是有上下级关系的路径,所以一个缩点之后一个简单的bfs就行了。

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef __int64 ll;
const int maxn=300100;
typedef pair<int,int> PI;
struct Edge{
    int from,next,to,num;
}e[maxn*2];
int tot,head[maxn];
int Index,top,DFN[maxn],Low[maxn],Stack[maxn],Belong[maxn];
int block;
bool Instack[maxn];
void init(){
    tot=0;
    mem1(head);
}

void addedge(int from,int to,int w){
    e[tot].from=from;
    e[tot].to=to;
    e[tot].next=head[from];
    e[tot].num=w;
    head[from]=tot++;
}

void Tarjan(int u,int pre){
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    int pre_cnt=0;
    for(int i=head[u];i!=-1;i=e[i].next){
        v=e[i].to;
        if(v==pre&&pre_cnt==0){
            pre_cnt++;
            continue;
        }
        if(!DFN[v]){
            Tarjan(v,u);
            if(Low[u]>Low[v])
                Low[u]=Low[v];
        }
        else if(Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if(Low[u]==DFN[u]){
        block++;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=block;
        }while(v!=u);
    }
}
int num[maxn],vis[maxn];

void bfs(int st){
    queue<int>Q;
    Q.push(st);
    vis[st]=1;
    while(!Q.empty()){
        int u=Q.front();
        Q.pop();
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].to,w=e[i].num;
            if(vis[v]==0){
                num[v]=max(num[v],num[u]+w);
                vis[v]=1;
                Q.push(v);
            }
        }
    }
}

void solve(int n,int st,int ed){
    mem0(DFN);
    mem0(vis);
    memset(Instack,false,sizeof(Instack));
    Index=top=block=0;
    for(int i=1;i<=n;i++)
        if(!DFN[i])
            Tarjan(i,0);
    int tot1=tot;
    init();
    for(int i=0;i<tot1;i++){
        int u=Belong[e[i].from],v=Belong[e[i].to];
        if(u==v)
            num[u]|=e[i].num;
        else
            addedge(u,v,e[i].num);
    }
    bfs(Belong[st]);
    if(vis[Belong[ed]]==1&&num[Belong[ed]]!=0)
        printf("YES\n");
    else
        printf("NO\n");
}

int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    int u,v,w;
    init();
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&u,&v,&w);
        addedge(u,v,w);
        addedge(v,u,w);
    }
    int st,ed;
    scanf("%d%d",&st,&ed);
    solve(n,st,ed);
    return 0;
}