今天学习了字典树
做了一些以前用二分搜索解决的题目
hdoj的1075
http://acm.hdu.edu.cn/showproblem.php?pid=1075用很脑残的方法过的，每个节点有设置11个字符(n)记录，这样浪费空间很大
看过后台数据，最大长度是4，有26^4个（字典序aaaa到zzzz）如果再多一点，内存就爆了
效率很低。。这道题目字典树不如二分
hdoj的1247
http://acm.hdu.edu.cn/showproblem.php?pid=1247也用了不是很好的方法过掉，枚举每个字符分成两段，看是不是在字典树内
用n标记是不是一个单词
hdoj的1251
http://acm.hdu.edu.cn/showproblem.php?pid=1251这个就是最基础的字典树了，n来标记这个前缀的次数
hdoj的1298
http://acm.hdu.edu.cn/showproblem.php?pid=1298这道就比前几道难了，dfs所有数字变成字母的可能情况，然后找到最大的概率的打印
dfs借鉴了yifeifen的int hash[10][2]={0,0, 0,0, 0,2, 3,5, 6,8, 9,11, 12,14, 15,18, 19, 21, 22,25};来记录每个数字对应的字母
这个就方便多了。。。
我用了最差的效率过的，每次都重新找一边，本来还超时的，后来优化一下，已经MANUALLY的就不用继续找下去了，后来的保证都MANUALLY
了。。。
还有usaco的一道
1 Contact
2 IOI'98
3
4 The cows have developed a new interest in scanning the universe outside their farm with radiotelescopes. Recently, they noticed a very curious microwave pulsing emission sent right from the centre of the galaxy. They wish to know if the emission is transmitted by some extraterrestrial form of intelligent life or if it is nothing but the usual heartbeat of the stars.
5
6 Help the cows to find the Truth by providing a tool to analyze bit patterns in the files they record. They are seeking bit patterns of length A through B inclusive (1 <= A <= B <= 12) that repeat themselves most often in each day's data file. They are looking for the patterns that repeat themselves most often. An input limit tells how many of the most frequent patterns to output.
7
8 Pattern occurrences may overlap, and only patterns that occur at least once are taken into account.
9 PROGRAM NAME: contact
10 INPUT FORMAT
11 Line 1: Three space-separated integers: A, B, N; (1 <= N < 50)
12 Lines 2 and beyond: A sequence of as many as 200,000 characters, all 0 or 1; the characters are presented 80 per line, except potentially the last line.
13 SAMPLE INPUT (file contact.in)
14
15 2 4 10
16 01010010010001000111101100001010011001111000010010011110010000000
17
18
19 In this example, pattern 100 occurs 12 times, and pattern 1000 occurs 5 times. The most frequent pattern is 00, with 23 occurrences.
20 OUTPUT FORMAT
21
22 Lines that list the N highest frequencies (in descending order of frequency) along with the patterns that occur in those frequencies. Order those patterns by shortest-to-longest and increasing binary number for those of the same frequency. If fewer than N highest frequencies are available, print only those that are.
23
24 Print the frequency alone by itself on a line. Then print the actual patterns space separated, six to a line (unless fewer than six remain).
25 SAMPLE OUTPUT (file contact.out)
26
27 23
28 00
29 15
30 01 10
31 12
32 100
33 11
34 11 000 001
35 10
36 010
37 8
38 0100
39 7
40 0010 1001
41 6
42 111 0000
43 5
44 011 110 1000
45 4
46 0001 0011 1100
47
大意就是叫你找出
01010010010001000111101100001010011001111000010010011110010000000
长度2于4之间的出现频率最高10个串
是TTBJ大大介绍我做的。。
写了一个半小时，终于把sample过掉了，叫TTBJ帮我提交，记过RE。。。。
他帮我检查了一下说别用指针，常常会出现错误的。。。
给了我一个用数组的模板
可惜我试了下好像C语言无法使用，只能割爱了。。。
今天下午用TTBJ的模板再次尝试好几遍只后
终于AC了，好开心阿
1 #include<stdio.h>
2 #include<string.h>
3 #include<stdlib.h>
4 struct tree{
5 char data;
6 int cnt,child[2];
7 void init(char c){
8 memset(child,0,sizeof(child));
9 data = c;
10 cnt = 0;
11 }
12 }T[1000001];
13 struct H{
14 int n;
15 char ch[13];
16 }hh[10000];
17 int L = 1;
18 int a,b,n,k,ans;
19 char ch[200001],temp[13];
20
21 void Ins(char *a,int k,int idx)
22 {
23 if(!a[k])
24 {
25 T[idx].cnt ++;
26 return ;
27 }
28 if(T[idx].child[ a[k]-'0' ])
29 Ins(a,k+1,T[idx].child[a[k]-'0']);
30 else
31 {
32 T[idx].child[a[k]-'0'] = ++L;
33 T[L].init(a[k]);
34 Ins(a,k+1,L);
35 }
36 }
37 void Creat()
38 {
39 int i,j,l,ii,len = strlen(ch);
40 char s[13];
41 for(i=0;ch[i];i++)
42 {
43 for(l=a,j=i,ii=0;l<=b;l++)
44 {
45 if(i+l>len)
46 break;
47 for(;j<i+l;j++,ii++)
48 {
49 s[ii] = ch[j];
50 }
51 s[ii] = 0;
52 Ins(s,0,1);
53 }
54 }
55 }
56 void find(char *a,int k,int idx)
57 {
58 if(!a[k])
59 {
60 ans = T[idx].cnt;
61 return ;
62 }
63 if(T[idx].child[a[k]-'0']==0)
64 return ;
65 find(a,k+1,T[idx].child[a[k]-'0']);
66 }
67
68 void dfs(int i)
69 {
70 if(i==b)
71 return ;
72 temp[i+1] = 0;
73
74 temp[i] = '0';
75 if(a<=i+1)
76 {
77 strcpy(hh[k].ch,temp);
78 ans = 0;
79 find(temp,0,1);
80 hh[k].n = ans;//符合条件，找
81 k++;
82 }
83 dfs(i+1);
84
85 temp[i+1] = 0;
86 temp[i] = '1';
87 if(a<=i+1)
88 {
89 strcpy(hh[k].ch,temp);
90 ans = 0;
91 find(temp,0,1);
92 hh[k].n = ans;
93 k++;
94 }
95 dfs(i+1);
96 }
97 int to_2(char *a,char *b)
98 {
99 int aa=0,bb=0;
100 for(;*a;a++)
101 aa = aa*2 + *a -'0';
102 for(;*b;b++)
103 bb = bb*2 + *b -'0';
104 return aa>bb?1:-1;
105 }
106 int cmp(const void *a,const void *b)
107 {
108 struct H * c = (struct H *)a;
109 struct H * d = (struct H *)b;
110 if(d->n == c->n)
111 {
112 if(strlen(c->ch) == strlen(d->ch))
113 return to_2(c->ch,d->ch);
114 return strlen(c->ch) - strlen(d->ch);
115 }
116 return d->n - c->n;
117 }
118 int main()
119 {
120 int i,count,sum,cnt;
121 while (scanf("%d%d%d",&a,&b,&n)==3)
122 {
123 if(a>b)
124 a^=b,b^=a,a^=b;
125 ch[0] = 0;
126 while(scanf("%s",temp)==1)
127 strcat(ch,temp);
128 Creat();//建树
129 k = 0;
130 dfs(0);//枚举所有点
131 qsort(hh,k,sizeof(hh[0]),cmp);
132 count = -1;
133 i = 0;
134 sum = -1;
135
136 while (1)//输出
137 {
138 if(hh[i].n!=sum)
139 {
140 if(i)
141 puts("");
142 count ++;
143 if(count == n)
144 break;
145 sum = hh[i].n;
146 printf("%d\n",hh[i].n);
147 printf("%s",hh[i].ch);
148 i++;
149 if(hh[i].n==0)
150 {
151 puts("");
152 break;
153 }
154 cnt = 1;
155 }
156 else
157 {
158 cnt ++;
159 if(cnt == 7)
160 {
161 cnt = 1;
162 puts("");
163 }
164 else
165 printf(" ");
166 printf("%s",hh[i].ch);
167 i++;
168 if(hh[i].n==0)
169 {
170 puts("");
171 break;
172 }
173 }
174 }
175 }
176 return 0;
177 }
好吧，字典树算是掌握了基础，明天的任务：
线段树~！
posted on 2009-02-10 02:11
shǎ崽 阅读(510)
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