• Iron and Coal(3次bfs)


    Iron and Coal(3次bfs)

     

     

    题意:给一个有向图,然后一些点有铁一些点有煤,然后问从1走,至少花费多少(走一个没走过的点花费1,走过的点再走一次是不用花费的)可以拿到一个铁一个煤,点可以重复走。

    思路:枚举从1节点到一个铁一个煤的交叉点即可,三遍bfs处理出dis[1,2,3][i]分别表示1号点到i的最小花费,i到铁的最小花费,i到煤的最小花费。

    AC_Code:

      1 #include <bits/stdc++.h>
      2 using namespace std;
      3 typedef long long ll;
      4 const int maxn = 1e5+10;
      5 const int inf = 0x3f3f3f3f;
      6 const double pi = acos(-1.0);
      7 #define rep(i,first,second) for(ll i=first;i<=second;i++)
      8 #define dep(i,first,second) for(ll i=first;i>=second;i--)
      9 
     10 ll n,m,k;
     11 ll dis[4][maxn],vis[maxn];
     12 vector<ll>g[maxn],ng[maxn];
     13 void bfs1(){
     14     queue<ll>q;
     15     while(!q.empty()) q.pop();
     16     rep(i,1,n+5) vis[i]=0;
     17     q.push(1);
     18     vis[1]=1;
     19     dis[1][1]=0;
     20     while( !q.empty() ){
     21         ll u=q.front(),v;
     22         q.pop();
     23         for(ll i=0;i<g[u].size();i++){
     24             v=g[u][i];
     25             if( vis[v] ) continue;
     26             vis[v]=1;
     27             dis[1][v]=dis[1][u]+1;
     28             q.push(v);
     29         }
     30     }
     31 }
     32 void bfs2(){
     33     queue<ll>q;
     34     while( !q.empty()) q.pop();
     35     rep(i,1,n+5) vis[i]=0;
     36     q.push(n+1);
     37     vis[n+1]=1;
     38     dis[2][n+1]=0;
     39     while( !q.empty() ){
     40         ll u=q.front(),v;
     41         q.pop();
     42         for(ll i=0;i<ng[u].size();i++){
     43             v=ng[u][i];
     44             if( vis[v] ) continue;
     45             if( u==n+1 ) dis[2][v]=dis[2][n+1];
     46             else dis[2][v]=dis[2][u]+1;
     47             vis[v]=1;
     48             q.push(v);
     49         }
     50     }
     51 }
     52 void bfs3(){
     53     queue<ll>q;
     54     while( !q.empty() ) q.pop();
     55     rep(i,1,n+5) vis[i]=0;
     56     q.push(n+2);
     57     vis[n+2]=1;
     58     dis[3][n+2]=0;
     59     while( !q.empty()){
     60         ll u=q.front(),v;
     61         q.pop();
     62         for(ll i=0;i<ng[u].size();i++){
     63             v=ng[u][i];
     64             if( vis[v]) continue;
     65             if( u==n+2 ) dis[3][v]=dis[3][n+2];
     66             else dis[3][v]=dis[3][u]+1;
     67             vis[v]=1;
     68             q.push(v);
     69         }
     70     }
     71 }
     72 
     73 int main()
     74 {
     75     scanf("%lld%lld%lld",&n,&m,&k);
     76     rep(i,1,n) dis[1][i]=dis[2][i]=dis[3][i]=inf;
     77     rep(i,1,m){
     78         ll o;
     79         scanf("%lld",&o);
     80         ng[n+1].push_back(o);
     81     }
     82     rep(i,1,k){
     83         ll c;
     84         scanf("%lld",&c);
     85         ng[n+2].push_back(c);
     86     }
     87     rep(i,1,n){
     88         ll tmp,v;
     89         scanf("%lld",&tmp);
     90         while( tmp-- ){
     91             scanf("%lld",&v);
     92             g[i].push_back(v);
     93             ng[v].push_back(i);
     94         }
     95     }
     96     bfs1();
     97     bfs2();
     98     bfs3();
     99     ll ans=inf;
    100     rep(i,1,n){
    101         ans=min(ans,dis[1][i]+dis[2][i]+dis[3][i]);
    102     }
    103     if( ans==inf ) puts("impossible");
    104     else{
    105         printf("%lld
    ",ans);
    106     }
    107     return 0;
    108 }
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  • 原文地址:https://www.cnblogs.com/wsy107316/p/12681158.html
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