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  • BNUOJ-26480 Horror List 最短路

      题目链接:http://www.bnuoj.com/bnuoj/problem_show.php?pid=26480

      题意:简单来说,就是给一个图,然后从每个honor list中的点求最短路。。

      边权值为1,Bfs就可以了,注意这里是无向图。。

      1 //STATUS:C++_AC_84MS_1788KB
      2 #include <functional>
      3 #include <algorithm>
      4 #include <iostream>
      5 //#include <ext/rope>
      6 #include <fstream>
      7 #include <sstream>
      8 #include <iomanip>
      9 #include <numeric>
     10 #include <cstring>
     11 #include <cassert>
     12 #include <cstdio>
     13 #include <string>
     14 #include <vector>
     15 #include <bitset>
     16 #include <queue>
     17 #include <stack>
     18 #include <cmath>
     19 #include <ctime>
     20 #include <list>
     21 #include <set>
     22 //#include <map>
     23 using namespace std;
     24 //#pragma comment(linker,"/STACK:102400000,102400000")
     25 //using namespace __gnu_cxx;
     26 //define
     27 #define pii pair<int,int>
     28 #define mem(a,b) memset(a,b,sizeof(a))
     29 #define lson l,mid,rt<<1
     30 #define rson mid+1,r,rt<<1|1
     31 #define PI acos(-1.0)
     32 //typedef
     33 typedef long long LL;
     34 typedef unsigned long long ULL;
     35 //const
     36 const int N=1010;
     37 const int INF=0x3f3f3f3f;
     38 const int MOD=1e9+7,STA=8000010;
     39 //const LL LNF=1LL<<60;
     40 const double EPS=1e-8;
     41 const double OO=1e15;
     42 const int dx[4]={-1,0,1,0};
     43 const int dy[4]={0,1,0,-1};
     44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
     45 //Daily Use ...
     46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
     47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
     48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
     49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
     50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
     51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
     52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
     53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
     54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
     55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
     56 //End
     57 
     58 struct Edge{
     59     int u,v;
     60 }e[N*10*2];
     61 int first[N],next[N*10*2],vis[N],hig[N];
     62 int n,m,L,mt;
     63 
     64 void adde(int a,int b)
     65 {
     66     e[mt].u=a,e[mt].v=b;
     67     next[mt]=first[a],first[a]=mt++;
     68     e[mt].u=b,e[mt].v=a;
     69     next[mt]=first[b],first[b]=mt++;
     70 }
     71 
     72 queue<int> q;
     73 void bfs()
     74 {
     75     int i,j,u,v;
     76     while(!q.empty())
     77     {
     78         u=q.front();q.pop();
     79         for(i=first[u];i!=-1;i=next[i]){
     80             v=e[i].v;
     81             if(vis[v])continue;
     82             vis[v]=1;
     83             hig[v]=hig[u]+1;
     84             q.push(v);
     85         }
     86     }
     87 }
     88 
     89 int main()
     90 {
     91  //   freopen("in.txt","r",stdin);
     92     int i,j,k,a,b,c;
     93     while(~scanf("%d%d%d",&n,&m,&L))
     94     {
     95         mem(first,-1);mt=0;
     96         mem(hig,INF);
     97         mem(vis,0);
     98         for(i=0;i<m;i++){
     99             scanf("%d",&a);
    100             hig[a]=0;
    101             q.push(a);
    102             vis[a]=1;
    103         }
    104         for(i=0;i<L;i++){
    105             scanf("%d%d",&a,&b);
    106             adde(b,a);
    107         }
    108 
    109         bfs();
    110         int ans=0,answ=0;
    111         for(i=0;i<n;i++){
    112             if(hig[i]>ans){
    113                 ans=hig[i];
    114                 answ=i;
    115             }
    116         }
    117 
    118         printf("%d
    ",answ);
    119     }
    120     return 0;
    121 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/3282042.html
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