C++博客-A Za, A Za, Fighting...-随笔分类-A_排序

PKU 2408 Anagram Groups (排序)

simplyzhao — Fri, 05 Nov 2010 07:38:00 GMT

Anagram Groups

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2318		Accepted: 649

Description

World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups.

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group. Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace(or new line). It is terminated by EOF. You can assume there will be no more than 30000 words.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

Source

Ulm Local 2000

思路:
这题将排序发挥到了极致啊呵呵，排序来排序去就AC了

代码:

1 /* 47MS */
2 #include<stdio.h>
3 #include<stdlib.h>
4 #include<string.h>
5 #define MAX_NUM 30001
6 #define MAX_LEN 36
7 #define MAX_OUT 5
8 struct Word {
9     char word[MAX_LEN];
10     char word_cmp[MAX_LEN];
11 } words[MAX_NUM];
12
13 struct Summary {
14     struct Word *first;
15     int count;
16 } smmry[MAX_NUM];
17
18 int total, total_category;
19
20 int
21 cmp_char(const void *arg1, const void *arg2)
22 {
23     return (*(char *)arg1) - (*(char *)arg2);
24 }
25
26 int
27 cmp_words(const void *arg1, const void *arg2)
28 {
29     int ret = strcmp(((struct Word *)arg1)->word_cmp, ((struct Word *)arg2)->word_cmp);
30     if(ret == 0)
31         ret = strcmp(((struct Word *)arg1)->word, ((struct Word *)arg2)->word);
32     return ret;
33 }
34
35 int
36 cmp_category(const void *arg1, const void *arg2)
37 {
38     int ret = ((struct Summary *)arg2)->count - ((struct Summary *)arg1)->count;
39     if(ret == 0)
40         ret = strcmp(((struct Summary *)arg1)->first->word, ((struct Summary *)arg2)->first->word);
41     return ret;
42 }
43
44 int
45 main(int argc, char **argv)
46 {
47     int i, j, num, len;
48     total = total_category = 0;
49     while(scanf("%s", words[total].word) != EOF) {
50         len = strlen(words[total].word);
51         strcpy(words[total].word_cmp, words[total].word);
52         qsort(words[total].word_cmp, len, sizeof(char), cmp_char);
53         ++total;
54     }
55     qsort(words, total, sizeof(struct Word), cmp_words);
56
57     num = 1;
58     for(i=1; i<total; i++) {
59         if(strcmp(words[i].word_cmp, words[i-1].word_cmp) == 0)
60             ++num;
61         else {
62             smmry[total_category].first = words+i-num;
63             smmry[total_category].count = num;
64             ++total_category;
65             num = 1;
66         }
67     }
68     smmry[total_category].first = words+i-num;
69     smmry[total_category++].count = num;
70     qsort(smmry, total_category, sizeof(struct Summary), cmp_category);
71
72     total_category = total_category < MAX_OUT ? total_category : MAX_OUT;
73     for(i=0; i<total_category; i++) {
74         printf("Group of size %d: %s ", smmry[i].count, smmry[i].first->word);
75         for(j=1; j<smmry[i].count; j++)
76             if(strcmp((smmry[i].first+j)->word, (smmry[i].first+j-1)->word) != 0)
77                 printf("%s ", (smmry[i].first+j)->word);
78         printf(".\n");
79     }
80 }

simplyzhao 2010-11-05 15:38 发表评论

PKU 2418 Hardwood Species

simplyzhao — Fri, 29 Oct 2010 16:59:00 GMT

问题:
http://poj.org/problem?id=2418

思路:
题意清晰明了，不难，用三种方法分别实现:
快速排序
动态生成节点的二叉查找树
静态分配节点的二叉查找树

结果发现，原来对于快速排序与静态分配节点都不是很熟悉，二维数组的快速排序分析见http://www.cppblog.com/Joe/archive/2010/10/29/131746.html，而动态生成节点则需要注意如果函数需要修改指针，那么必须传递指向指针的指针，因为C是值传递的

另外，我以为静态分配节点应该比动态分配节点节约很多时间的，结果居然差不多...而快速排序在这题明显是最耗时的

代码(快排)

1 /* 35640K 1844MS */
2 #include<stdio.h>
3 #include<stdlib.h>
4 #include<string.h>
5 #define MAX_LEN 36
6 #define MAX_NUM 1000001
7 char species[MAX_NUM][MAX_LEN];
8
9 int
10 cmp(const void *arg1, const void *arg2)
11 {
12     return strcmp((char *)arg1, (char *)arg2);
13 }
14
15 int
16 main(int argc, char **argv)
17 {
18     int i, count, total = 0;
19     while(gets(species[total]) != NULL)
20         ++total;
21     qsort(species, total, sizeof(species[0]), cmp);
22     count = 1;
23     for(i=1; i<total; i++) {
24         if(strcmp(species[i], species[i-1]) == 0)
25             ++count;
26         else {
27             printf("%s %.4f\n", species[i-1], (count*100.0)/total);
28             count = 1;
29         }
30     }
31     printf("%s %.4f\n", species[total-1], (count*100.0)/total);
32 }

代码(动态分配节点的BST，原本想实现下destroy函数的，结果怕麻烦就留给系统回收吧(*^__^*) 嘻嘻……)

1 /* binary search tree(dynamic allocation) */
2 /* 544K 1188MS */
3 #include<stdio.h>
4 #include<stdlib.h>
5 #include<string.h>
6 #include<assert.h>
7 #define MAX_LEN 36
8 struct Node {
9     char spec[MAX_LEN];
10     int count;
11     struct Node *left, *right;
12 };
13 int total;
14
15 struct Node *
16 create_node(char *str)
17 {
18     struct Node *node = (struct Node *)malloc(sizeof(struct Node));
19     assert(node != NULL);
20     strcpy(node->spec, str);
21     node->left = node->right = NULL;
22     node->count = 1;
23     return node;
24 }
25
26 void
27 insert(struct Node **root, char *str)
28 {
29     int ret;
30     struct Node *node;
31     if(*root==NULL) {
32         *root = create_node(str);
33         return;
34     }
35     ret = strcmp((*root)->spec, str);
36     if(ret == 0)
37         ++((*root)->count);
38     else if(ret < 0)
39         insert(&((*root)->right), str);
40     else
41         insert(&((*root)->left), str);
42 }
43
44 void
45 inorder(struct Node *root)
46 {
47     if(root == NULL)
48         return;
49     inorder(root->left);
50     printf("%s %.4f\n", root->spec, (root->count)*100.0/total);
51     inorder(root->right);
52 }
53
54 void
55 destroy(struct Node **root)
56 {
57 }
58
59 int
60 main(int argc, char **argv)
61 {
62     char str[MAX_LEN];
63     struct Node *bst = NULL;
64     total = 0;
65     while(gets(str) != NULL) {
66         ++total;
67         insert(&bst, str);
68     }
69     inorder(bst);
70 }

代码(静态分配节点的BST)

1 /* binary search tree(static allocation) */
2 /* 492K 1188MS */
3 #include<stdio.h>
4 #include<stdlib.h>
5 #include<string.h>
6 #include<assert.h>
7 #define MAX_LEN 36
8 #define MAX_NUM 10007
9 #define ROOT 1
10 struct Node {
11     char spec[MAX_LEN];
12     int count;
13     int left, right;
14 }bst[MAX_NUM];
15 int cur_index, total;
16
17 int
18 find(int root, char *str)
19 {
20     int ret, parent, cur = root;
21     while(cur != 0) {
22         parent = cur;
23         ret = strcmp(bst[cur].spec, str);
24         if(ret == 0) {
25             ++bst[cur].count;
26             return 0;
27         } else if(ret < 0) {
28             cur = bst[cur].right;
29         } else {
30             cur = bst[cur].left;
31         }
32     }
33     return parent;
34 }
35
36 #define ADD(index, str) { \
37     strcpy(bst[index].spec, str); \
38     bst[index].left = bst[index].right = 0; \
39     bst[index].count = 1; \
40     ++index; }
41
42 void
43 insert(int parent, char *str)
44 {
45     int ret = strcmp(bst[parent].spec, str);
46     assert(ret != 0);
47     if(ret < 0)
48         bst[parent].right = cur_index;
49     else
50         bst[parent].left = cur_index;
51     ADD(cur_index, str);
52 }
53
54 void
55 inorder(int index)
56 {
57     if(index == 0)
58         return;
59     inorder(bst[index].left);
60     printf("%s %.4f\n", bst[index].spec, (bst[index].count*100.0)/total);
61     inorder(bst[index].right);
62 }
63
64 int
65 main(int argc, char **argv)
66 {
67     int parent;
68     char str[MAX_LEN];
69     total = 1;
70     cur_index = ROOT;
71     gets(str);
72     ADD(cur_index, str); /* create the root node first */
73     while(gets(str) != NULL) {
74         ++total;
75         if((parent=find(ROOT, str)) > 0)
76             insert(parent, str);
77     }
78     inorder(ROOT);
79 }

simplyzhao 2010-10-30 00:59 发表评论

[Tips][Original] qsort应用于指针数组与二维数组(字符)的差异

simplyzhao — Fri, 29 Oct 2010 07:09:00 GMT

在将qsort函数应用于对指针数组与二维数组排序时，传递给compare函数的参数类型是不同的

首先，我们举个简单的例子，先将qsort对整数数组排序:

1 int
2 cmp(const void *arg1, const void *arg2)
3 {
4     return (*(int *)arg1)-(*(int *)arg2);
5 }
6
7 int
8 main(int argc, char **argv)
9 {
10     int i;
11     int arr[] = {3, 1, 5, 2, 4};
12     qsort(arr, sizeof(arr)/sizeof(arr[0]), sizeof(int), cmp);
13 }

排序针对的是数组里的元素而言的，这里整数数组的元素就是整数，因此qsort的第三个参数就是sizeof(int)，而传递给比较函数cmp的参数就是相对应的指向整数的指针

接着，我们来看看指针数组的情形:

1 int
2 cmp(const void *arg1, const void *arg2)
3 {
4     return strcmp((*(char **)arg1), (*(char **)arg2));
5 }
6
7 int
8 main(int argc, char **argv)
9 {
10     int i;
11     /* pointer array */
12     char *str[] = {"java", "c", "python", "perl"};
13     qsort(str, sizeof(str)/sizeof(str[0]), sizeof(char *), cmp);
14 }

这里的理解其实跟整数数组差不多，关键是抓住数组里元素的类型，既然称之为指针数组，那数组元素的类型自然就是指针，因此qsort的第三个参数就是sizeof(char *)，而传递给比较函数cmp的参数就是相对应的指向指针的指针，这里即char **类型

二维数组的理解最为复杂，代码如下:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4
5 int
6 cmp1(const void *arg1, const void *arg2)
7 {
8     return strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2)));
9 }
10
11 int
12 cmp2(const void *arg1, const void *arg2)
13 {
14     return strcmp((char *)arg1, (char *)arg2);
15 }
16
17 int
18 main(int argc, char **argv)
19 {
20     int i;
21     char str1[4][8] = {"java", "c", "python", "peal"};
22     printf("COMPARE-FUNCTION-1\n");
23     qsort(str1, 4, sizeof(str1[0]), cmp1);
26
27     char str2[4][8] = {"java", "c", "python", "peal"};
28     printf("COMPARE-FUNCTION-2\n");
29     qsort(str2, 4, sizeof(str2[0]), cmp2);
34 }

这里cmp1与cmp2都能正常的工作(*^__^*) 嘻嘻……
还是按照上述方法来分析，抓住数组元素的类型来入手，二维数组实际上就是数组的数组，因此二维数组的元素类型就是一维数组，因此qsort的第三个参数就是sizeof(str1[0])或sizeof(str2[0])，那传递给比较函数的参数应该就是指向数组的指针，这点可以通过gdb设置断点来得到证实:

1 (gdb) p &str1[0]
2 $1 = (char (*)[8]) 0xbffff2cc
3 (gdb) p &str1[1]
4 $2 = (char (*)[8]) 0xbffff2d4
5
6 Breakpoint 2, cmp1 (arg1=0xbffff2cc, arg2=0xbffff2d4) at char_test2.c:8
7 8 return strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2)));
8 (gdb) p arg1
9 $3 = (const void *) 0xbffff2cc
10 (gdb) p arg2
11 $4 = (const void *) 0xbffff2d4
12 (gdb) p *(char (*)[])arg1
13 $5 = "j"
14 (gdb) p *(char (*)[8])arg1
15 $6 = "java\000\000\000"

通过第2行与第9行的比较可以发现，比较函数的参数arg1其实就是&str1[0]，类型为char (*)[]，这也是为什么cmp1能正常工作的原因
那么cmp2呢，它为什么正确呢?
在cmp1中：strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2))); 这里传递给strcmp的参数之所以不会出错，是因为我们将arg1解地址操作获得一个数组，而数组名其实是指向数组首元素的指针，arg1既然是指向str1[0]这个一维数组的指针，而str1[0]本身其实就是指向这个一维数组的指针，也就是说arg1其实就是str1[0]，因此cmp2能够正常工作

1 (gdb) p &str1[0]
2 $3 = (char (*)[8]) 0xbffff2cc
3 (gdb) p &str1[0][0]
4 $4 = 0xbffff2cc "java"
5 (gdb) p arg1
6 $5 = (const void *) 0xbffff2cc
7 (gdb) p (char *)arg1
8 $6 = 0xbffff2cc "java"

额...貌似越说越复杂的样子，不过这是我理解的过程，见谅...

simplyzhao 2010-10-29 15:09 发表评论

PKU 1089 Intervals

simplyzhao — Sun, 17 Oct 2010 11:35:00 GMT

问题:
http://poj.org/problem?id=1089

思路:
根据interval的begin升序排列，然后对于interval A, B只存在三种情况:
A包含B、A与B相交、A与B分离

代码:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #define MAX_LEN 50001
5 struct Interval {
6     int begin, end;
7 }itvs[MAX_LEN], target[MAX_LEN];
8 int N, total;
9
10 int
11 compare(const void *arg1, const void *arg2)
12 {
13     struct Interval *a = (struct Interval *)arg1;
14     struct Interval *b = (struct Interval *)arg2;
15     if(a->begin == b->begin)
16         return a->end - b->end;
17     return a->begin - b->begin;
18 }
19
20 void
21 init()
22 {
23     int i;
24     for(i=0; i<N; i++)
25         scanf("%d %d", &itvs[i].begin, &itvs[i].end);
26     qsort(itvs, N, sizeof(struct Interval), compare);
27     total = 0;
28 }
29
30 void
31 solve()
32 {
33     int i;
34     struct Interval cur = itvs[0];
35     for(i=1; i<N; i++) {
36         if(itvs[i].begin > cur.end) {
37             target[total++] = cur;
38             cur = itvs[i];
39         } else {
40             if(itvs[i].end > cur.end)
41                 cur.end = itvs[i].end;
42         }
43     }
44     target[total++] = cur;
45 }
46
47 void
48 output()
49 {
50     int i;
51     for(i=0; i<total; i++)
52         printf("%d %d\n", target[i].begin, target[i].end);
53 }
54
55 int
56 main(int argc, char **argv)
57 {
58     while(scanf("%d", &N) != EOF) {
59         init();
60         solve();
61         output();
62     }
63 }

simplyzhao 2010-10-17 19:35 发表评论

PKU 1828 Monkeys' Pride

simplyzhao — Wed, 08 Sep 2010 15:41:00 GMT

问题:
http://acm.pku.edu.cn/JudgeOnline/problem?id=1828

思路:
按照坐标从上到下、从左到右排序
首先想到的是O(n^2)的算法，时间需要1600+MS
然后，发现其实在排序之后只要从后向前扫描一遍即可得出结果

代码:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #define MAX_LEN 50001
5 int n;
6 struct Point {
7     int x, y;
8 }points[MAX_LEN];
9
10 int
11 compare(const void *arg1, const void *arg2)
12 {
13     struct Point *a = (struct Point *)arg1;
14     struct Point *b = (struct Point *)arg2;
15     if(a->x == b->x)
16         return a->y - b->y;
17     return a->x - b->x;
18 }
19
20 int
21 main(int argc, char **argv)
22 {
23     int i, j, cnt, ymax;
24     while(scanf("%d", &n)!=EOF && n) {
25         for(i=0; i<n; i++)
26             scanf("%d %d", &points[i].x, &points[i].y);
27         qsort(points, n, sizeof(struct Point), compare);
28         /* O(n^2) AC 1600+MS
29         cnt = 0;
30         for(i=0; i31             for(j=i+1; j32                 if(points[j].y >= points[i].y)
33                     break;
34             }
35             if(j == n)
36                 ++cnt;
37         }
38         */
39         /* O(nlgn) AC 235MS */
40         cnt = 1;
41         ymax = points[n-1].y;
42         for(i=n-2; i>=0; i--) {
43             if(ymax < points[i].y) {
44                 ++cnt;
45                 ymax = points[i].y;
46             }
47         }
48         printf("%d\n", cnt);
49     }
50 }

simplyzhao 2010-09-08 23:41 发表评论

PKU 2388 Who's in the Middle

simplyzhao — Sun, 04 Jul 2010 02:19:00 GMT

问题:
http://acm.pku.edu.cn/JudgeOnline/problem?id=2388

思路:
1. 排序
这题直接用快排就能AC，很水...

2. 求解第i个顺序统计量(详见CLRS)
利用快速排序过程中的思路，期望时间复杂度是O(n)

ps: partition函数的执行过程是比较有意思的呵呵

1 int
2 partition(long *arr, int begin, int end)
3 {
4     int i, j;
5     i = begin;
6     long pivot = arr[begin];
7     for(j=begin+1; j<=end; j++)
8         if(arr[j] <= pivot)
9             swap(arr, ++i, j);
10     swap(arr, i, begin);
11     return i;
12 }
13
14 long
15 find_ith_os(long *arr, int begin, int end, int ith)
16 {
17     if(begin == end)
18         return arr[begin];
19     int p = partition(arr, begin, end);
20     int k = p-begin+1;
21     if(k == ith)
22         return arr[p];
23     else if(ith < k)
24         return find_ith_os(arr, begin, p-1, ith);
25     else
26         return find_ith_os(arr, p+1, end, ith-k);
27 }

simplyzhao 2010-07-04 10:19 发表评论

PKU 1007 DNA Sorting/2299 Ultra-QuickSort

simplyzhao — Sun, 04 Jul 2010 02:07:00 GMT

问题:
http://acm.pku.edu.cn/JudgeOnline/problem?id=1007
http://acm.pku.edu.cn/JudgeOnline/problem?id=2299

思路:
求逆序对的个数
这两题的基本问题是一致的，给定一个数组(包括字符串)，求出逆序对的个数

1. 最简单的方法
两层循环，复杂度O(n^2)

1 int
2 inversion_cal(char *str)
3 {
4     int i, j, count = 0;
5     int len = strlen(str);
6     for(i=0; i<len; i++)
7         for(j=i+1; j<len; j++)
8             if(str[i] > str[j])
9                 ++count;
10     return count;
11 }

2. 归并排序＆分治思想
其实，只要将归并排序稍加修改，就是一个求解逆序对个数问题的O(nlgn)方法
要理解的是这其中涉及的分治思想(三步骤):
a. 分解为子问题
b. 求解子问题
c. 合并子问题的解来得到原问题的解
具体对应到求逆序对个数的问题:
a. 将原数组分解为前后两个子数组
b. 求解子数组的逆序对个数
c. 合并前后子数组，同时计算逆序对个数(这个是指逆序对的第一个数在前子数组中，而第二个数在后子数组中)

1 long long
2 merge_count(long *arr, long *temp, long begin, long end)
3 {
4     if(begin >= end)
5         return 0;
6     long i, j, k, mid = (begin+end)/2;
7     long long rt = 0;
8     rt += merge_count(arr, temp, begin, mid);
9     rt += merge_count(arr, temp, mid+1, end);
10     i = k = begin;
11     j = mid+1;
12     while(i<=mid && j<=end) {
13         if(arr[i] <= arr[j])
14             temp[k++] = arr[i++];
15         else {
16             temp[k++] = arr[j++];
17             rt += (mid-i+1);
18         }
19     }
20     for( ; i<=mid; i++)
21         temp[k++] = arr[i];
22     for( ; j<=end; j++) {
23         temp[k++] = arr[j];
24         rt += (mid-i+1);
25     }
26     /* copy */
27     for(k=begin; k<=end; k++)
28         arr[k] = temp[k];
29     return rt;
30 }

3. 特例方法
针对PKU 1007该题的特殊性: 字符串中只包含A, G, T, C四个字母，还有一种更加简单的O(n)方法

1 int
2 inversion_cal2(char *str)
3 {
4     int i, temp[4], count = 0;
5     int len = strlen(str);
6     memset(temp, 0, sizeof(temp));
7     for(i=len-1; i>=0; i--) {
8         switch(str[i]) {
9             case 'A':
10                 ++temp[0];
11                 break;
12             case 'C':
13                 ++temp[1];
14                 count += temp[0];
15                 break;
16             case 'G':
17                 ++temp[2];
18                 count += (temp[0]+temp[1]);
19                 break;
20             case 'T':
21                 ++temp[3];
22                 count += (temp[0]+temp[1]+temp[2]);
23                 break;
24         }
25     }
26     return count;
27 }

simplyzhao 2010-07-04 10:07 发表评论