HDU-4547 CD操作 LCA
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4547
去年区预赛网络赛的时候写过一到LCA的题目<HDU4409>，本来想用那份代码修改下来交这道题的，不过貌似难得修改，就重新写了ST+DFS的代码当模板用。跑了2078ms，应该是map的效率低了。。。
1 //STATUS:C++_AC_2078MS_26116KB 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 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef __int64 LL; 33 typedef unsigned __int64 ULL; 34 //const 35 const int N= 200010; 36 const int INF=0x3f3f3f3f; 37 const int MOD=100000,STA=8000010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 struct Edge{ 58 int u,v; 59 }e[N]; 60 int first[N],next[N],f[N][20],d[N],E[N],R[N],vis[N]; 61 int T,n,m,mt,all,dis,root; 62 map<string,int> q; 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 } 69 70 int Minele(int i,int j) 71 { 72 return d[i]<d[j]?i:j; 73 } 74 75 void rmq_init(int n) 76 { 77 int i,j; 78 for(i=1;i<=n;i++)f[i][0]=i; 79 for(j=1;(1<<j)<=n;j++){ 80 for(i=1;i+(1<<j)-1<=n;i++){ 81 f[i][j]=Minele(f[i][j-1],f[i+(1<<(j-1))][j-1]); 82 } 83 } 84 } 85 86 int rmq(int l,int r) 87 { 88 int k=0; 89 while((1<<(k+1))<=r-l+1)k++; 90 return Minele(f[l][k],f[r-(1<<k)+1][k]); 91 } 92 93 void dfs(int u,int deep) 94 { 95 d[dis]=deep; 96 E[dis]=u; 97 R[u]=dis++; 98 int i; 99 for(i=first[u];i!=-1;i=next[i]){ 100 dfs(e[i].v,deep+1); 101 d[dis]=deep; 102 E[dis++]=u; 103 } 104 } 105 106 void LCA_Init() 107 { 108 dis=1; 109 dfs(root,0); 110 rmq_init(dis-1); 111 } 112 113 int LCA(int a,int b) 114 { 115 int left=R[a],right=R[b]; 116 if(left>right)swap(left,right); 117 return E[rmq(left,right)]; 118 } 119 120 int main() 121 { 122 // freopen("in.txt","r",stdin); 123 int i,j,fa,a,b; 124 char s1[42],s2[42]; 125 scanf("%d",&T); 126 while(T--) 127 { 128 scanf("%d%d",&n,&m); 129 q.clear();all=1; 130 mem(first,-1);mt=0; 131 mem(vis,0); 132 while(--n){ 133 scanf("%s%s",s1,s2); 134 if(!q[s1])q[s1]=all++; 135 if(!q[s2])q[s2]=all++; 136 vis[q[s1]]=1; 137 adde(q[s2],q[s1]); 138 } 139 for(i=1;i<all;i++)if(!vis[i]){root=i;break;} 140 141 LCA_Init(); 142 while(m--){ 143 scanf("%s%s",s1,s2); 144 a=q[s1];b=q[s2]; 145 fa=LCA(a,b); 146 printf("%d\n",d[R[a]]-d[R[fa]]+(d[R[b]]>d[R[fa]])); 147 } 148 } 149 return 0; 150 }