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\n",answ); 119 } 120 return 0; 121 }
