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。

/* ***********************************************
Author :111qqz
Created Time :2016年05月19日 星期四 15时44分12秒
File Name :code/poj/1470.cpp
************************************************ */
 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=905;
 7int n;
 8vector <int> edge[N];
 9int in[N];
10int E[2*N],R[2*N];
11int depth[2*N];
12int p;
13int dp[2*N][12];
14int cnt[N];
15void dfs( int u,int dep)
16{
17    p++;
18    E[p] = u ;
19    depth[p] =  dep;
20    R[u] = p;
21    int siz = edge[u].size();
22    for ( int i = 0 ; i < siz ; i ++)
23    {
24	int v  = edge[u][i];
1	dfs(v,dep+1);
2	p++;
3	E[p] =  u;
4	depth[p] = dep;
5    }
6}
1int _min( int l,int r)
2{
3    if (depth[l]<depth[r]) return l;
4    return r;
5}
6void rmq_init()
7{
8    for ( int i = 1 ; i <=2*n+2 ; i++) dp[i][0] =  i;
1    for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
2	for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
3	    dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
4}
 1int rmq_min( int l,int r)
 2{
 3    if (l>r) swap(l,r);
 4    int k = 0;
 5    while (1<<(k+1) <= r-l+1) k++;
 6    return _min(dp[l][k],dp[r-(1<<k)+1][k]);
 7}
 8int main()
 9{
10	#ifndef  ONLINE_JUDGE 
11	freopen("code/in.txt","r",stdin);
12  #endif
 1	while (scanf("%d",&n)!=EOF)
 2	{
 3	    ms(in,0);
 4	    for ( int i = 1 ; i <= n ; i++) edge[i].clear();
 5	    for ( int i = 1 ; i <= n ; i++)
 6	    {
 7		char ch[5];
 8		int x,num;
 9		scanf("%d%2s%d%1s",&x,ch,&num,ch);
10//		cout<<"x:"<<x<<" num:"<<num<<endl;
1		for ( int i = 1 ; i <= num ; i++)
2		{
3		    int y;
4		    scanf("%d",&y);
5		    edge[x].push_back(y);
6		    in[y]++;
7//		    cout<<"y:"<<y<<endl;
8		}
9	    }
1	    int root ;
2	    for ( int i = 1 ; i <= n ; i++) if (in[i]==0) root = i ;
3//	    cout<<"root:"<<root<<endl;
4	    p = 0;
5	    dfs(root,0);
6	    rmq_init();
	    ms(cnt,0);
 1	    int q;
 2	    scanf("%d",&q);
 3//	    cout<<"q:"<<q<<endl;
 4	    while (q--)
 5	    {
 6		char ch[5];
 7		int x,y;
 8		scanf("%1s%d%d%1s",ch,&x,&y,ch);
 9		int LCA = E[rmq_min(R[x],R[y])];
10//		cout<<"x:"<<x<<" y:"<<y<<" LCA:"<<LCA<<endl;
11		cnt[LCA]++;
12	    }
13//	    cout<<"n:"<<n<<endl;
14	    for ( int i = 1 ; i <= n ; i++)
15	    {
16//		cout<<"cnt[i]:"<<cnt[i]<<endl;
17		if (cnt[i]==0) continue;
18		printf("%d:%d\n",i,cnt[i]);
19	    }
20	}
1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}