poj 1470 Closest Common Ancestors (lca,rmq+dfs,读入技巧)
题意:求两点的lca. 思路:dfs+rmq. 读入技巧。 读入比较坑爹。。。 学会了一种新的读入技巧。
scanf("%2s",st);
表示读一个长度为2的字符串。。。读的时候会忽略各种空白字符。
1/* ***********************************************
2Author :111qqz
3Created Time :2016年05月19日 星期四 15时44分12秒
4File Name :code/poj/1470.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=905;
34int n;
35vector <int> edge[N];
36int in[N];
37int E[2*N],R[2*N];
38int depth[2*N];
39int p;
40int dp[2*N][12];
41int cnt[N];
42void dfs( int u,int dep)
43{
44 p++;
45 E[p] = u ;
46 depth[p] = dep;
47 R[u] = p;
48 int siz = edge[u].size();
49 for ( int i = 0 ; i < siz ; i ++)
50 {
51 int v = edge[u][i];
52
53 dfs(v,dep+1);
54 p++;
55 E[p] = u;
56 depth[p] = dep;
57 }
58}
59
60int _min( int l,int r)
61{
62 if (depth[l]<depth[r]) return l;
63 return r;
64}
65void rmq_init()
66{
67 for ( int i = 1 ; i <=2*n+2 ; i++) dp[i][0] = i;
68
69 for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
70 for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
71 dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
72}
73
74int rmq_min( int l,int r)
75{
76 if (l>r) swap(l,r);
77 int k = 0;
78 while (1<<(k+1) <= r-l+1) k++;
79 return _min(dp[l][k],dp[r-(1<<k)+1][k]);
80}
81int main()
82{
83 #ifndef ONLINE_JUDGE
84 freopen("code/in.txt","r",stdin);
85 #endif
86
87 while (scanf("%d",&n)!=EOF)
88 {
89 ms(in,0);
90 for ( int i = 1 ; i <= n ; i++) edge[i].clear();
91 for ( int i = 1 ; i <= n ; i++)
92 {
93 char ch[5];
94 int x,num;
95 scanf("%d%2s%d%1s",&x,ch,&num,ch);
96// cout<<"x:"<<x<<" num:"<<num<<endl;
97
98 for ( int i = 1 ; i <= num ; i++)
99 {
100 int y;
101 scanf("%d",&y);
102 edge[x].push_back(y);
103 in[y]++;
104// cout<<"y:"<<y<<endl;
105 }
106 }
107
108 int root ;
109 for ( int i = 1 ; i <= n ; i++) if (in[i]==0) root = i ;
110// cout<<"root:"<<root<<endl;
111 p = 0;
112 dfs(root,0);
113 rmq_init();
114
115 ms(cnt,0);
116
117 int q;
118 scanf("%d",&q);
119// cout<<"q:"<<q<<endl;
120 while (q--)
121 {
122 char ch[5];
123 int x,y;
124 scanf("%1s%d%d%1s",ch,&x,&y,ch);
125 int LCA = E[rmq_min(R[x],R[y])];
126// cout<<"x:"<<x<<" y:"<<y<<" LCA:"<<LCA<<endl;
127 cnt[LCA]++;
128 }
129// cout<<"n:"<<n<<endl;
130 for ( int i = 1 ; i <= n ; i++)
131 {
132// cout<<"cnt[i]:"<<cnt[i]<<endl;
133 if (cnt[i]==0) continue;
134 printf("%d:%d\n",i,cnt[i]);
135 }
136 }
137
138 #ifndef ONLINE_JUDGE
139 fclose(stdin);
140 #endif
141 return 0;
142}