poj 1470 Closest Common Ancestors (lca,rmq+dfs,读入技巧)

poj1470题目链接

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}