UVa 642 - Word Amalgamation

题目：给你一个单词列表。再给你一些新的单词。输出列表中又一次排列能得到此新单词的词。

分析：字符串。对每一个字符串的字母排序生成新的传f（str）。总体排序，用二分来查找就可以。

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;

}