UVa 642 - Word Amalgamation

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题目：给你一个单词列表，再给你一些新的单词，输出列表中重新排列能得到此新单词的词。

分析：字符串。对每个字符串的字母排序生成新的传f（str），整体排序，用二分来查找即可。

说明：注意输出要满足字典序，先排序后查找。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;typedef struct wnode{char word[7];char abcd[7];}words;words W[101];int cmp(words a, words b){int c = strcmp(a.abcd, b.abcd);if (c != 0) return c<0;return strcmp(a.word, b.word)<0;}int bs(char str[], int r){int l = 0,m,c;while (l < r) {m = (l+r)/2;c = strcmp(W[m].abcd, str);if (c < 0) l = m+1;else r = m;}return l;}char buf[7];int main(){int count = 0;while (gets(W[count].word) && strcmp(W[count].word, "XXXXXX")) {strcpy(W[count].abcd, W[count].word);sort(W[count].abcd, W[count].abcd+strlen(W[count].abcd));count ++;}sort(W, W+count, cmp);while (gets(buf) && strcmp(buf, "XXXXXX")) {sort(buf, buf+strlen(buf));int s = bs(buf, count-1),flag = 0;while (!strcmp(buf, W[s].abcd)) {printf("%s\n",W[s ++].word);flag = 1;}if (!flag) printf("NOT A VALID WORD\n");printf("******\n");}return 0;}

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