poj2945

前阵子用Trie树过了poj2945，当然题目本身比较水，但数据量应该还是比较大的。限制时间为5秒。

4548K && 547MS

Trie树解法：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int const childnum = 4;const int map[] = {0, -1, 1, -1, -1, -1, 2,-1, -1, -1, -1, -1, -1,-1, -1, -1, -1, -1, -1, 3};struct node {node* child[childnum];int times;bool isstop;};const int peoplenum = 20010;const int DNAlen = 21;node Tree[peoplenum*DNAlen];int nodecnt = 0;//rec[i] contains the number of persons that are present in i identical copies//rec[4] = 3 重复了4次的不同基因有4个int rec[peoplenum];char str[peoplenum][DNAlen];void init(node *p){memset(p->child, NULL, sizeof (p->child));p->isstop = false;p->times = 0;}void insert(node *p, char *s){int rec = nodecnt;for (int i = 0; s[i]; i++) {int idx = map[s[i]-'A'];if (p->child[idx] == NULL) {nodecnt++;p->child[idx] = Tree + nodecnt;init(p->child[idx]);}p = p->child[idx];}if (p->isstop == true) {(p->times++);return;}p->isstop = true;p->times = 1;}int main(){int n, m;while (scanf("%d%d", &n, &m), n || m) {node *p = Tree;init(p);nodecnt = 0;for (int i = 0; i != n; i++) {scanf("%s", str[i]);insert(p, str[i]);}memset(rec, 0, sizeof (rec));for (int i = 1; i != nodecnt+1; i++) {if (Tree[i].isstop = true) {rec[Tree[i].times]++;}}for (int i = 1; i != n+1; i++) {printf("%d\n", rec[i]);}}return 0;}

排序解法

最近对STL比较有兴趣，用vector<string>来解，其实就是字符串数组，结果超时了。。。不过考虑用scanf()和printf()改改再试。

结果马上换成C语言的东西来写，AC了。

1244K && 985MS

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node {char DNA[25];};node str[20010];int counter[20010];bool cmp(node a, node b){for (int i = 0; i != strlen(a.DNA); i++) {if (a.DNA[i] != b.DNA[i])return a.DNA[i] < b.DNA[i];}return true;}int main(){int cnt, len, i;while (scanf("%d%d", &cnt, &len) != EOF && cnt || len) {for (i = 0; i != cnt; i++) {counter[i] = 0;memset(str[i].DNA, 0, sizeof (str[i].DNA));}for (i = 0; i != cnt; i++) scanf("%s", str[i].DNA);sort(str, str+cnt, cmp);for (i = 0; i != len; i++) {str[cnt].DNA[i] = 'Z';}str[cnt].DNA[i] = '\0';int num = 0;for (i = 1; i != cnt+1; i++) {if (strcmp(str[i].DNA, str[i-1].DNA) == 0) {num++;}else {counter[num+1]++;num = 0;}}for (i = 1; i != cnt+1; i++) {printf("%d\n", counter[i]);}} //whilereturn 0;}

今天学了一下STL的map，使用之，写起代码来，很短很清晰！当然了，回防止超时，还是用了C语言的字符串和C的输入输出。

1892K && 1532MS

#include <iostream>#include <cstdio>#include <string>#include <map>using namespace std;const int people = 20010;const int DNAlen = 25;typedef map<string, int> msi;int cnt[people];int main(){msi mapsi;int num, len;char tmp[DNAlen];while (cin >> num >> len, num||len) {mapsi.clear();for (int i = 1; i != num+1; i++)cnt[i] = 0;for (int i = 0; i != num; i++) {scanf("%s", tmp);mapsi[tmp]++;}for (msi::iterator it = mapsi.begin(); it != mapsi.end(); it++)cnt[it->second]++;for (int i = 1; i != num+1; i++)printf("%d\n", cnt[i]);}return 0;}