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UVa 10100 - Longest Match

题目:求两组字符串中最大的按顺序出现的相同单词数目。

分析:dp,最大公共子序列(LCS)。把单词整个看成一个元素比较即可。

            状态:f(i,j)为s1串前i个单词与s2串前j个单词的最大匹配数;

            转移:f(i,j)= max(f(i-1,j),f(i,j-1)){ s1[i] ≠ s2[j] };

                                           f(i-1,j-1)+ 1。

            这里的字母包括数字

说明:数据输入,需要先分解成单词,然后计算即可。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

char buf[1001]; 
char s1[1001][22];
char s2[1001][22];

int  f[1001][1001];

int letter(char c)
{
	if (c >= 'a' && c <= 'z')
		return 1;
	if (c >= 'A' && c <= 'Z')
		return 1;
	if (c >= '0' && c <= '9') 
		return 1;
	return 0;
}

int getword(char s[][22], char *t)
{
	int move = 0,count = 0;
	while (t[move]) {
		while (t[move] && !letter(t[move])) 
			move ++;
		if (!t[move]) break;
		int now = move;
		while (letter(t[move])) 
			move ++;
		int save = 0;
		while (now < move) 
			s[count][save ++] = t[now ++];
		s[count][save] = 0;
		count ++;
	}
	return count;
}

int main()
{
	int blank = 0,l1,l2,cases = 1;
	while (gets(buf)) {
		if (!strlen(buf)) blank = 1;
		l1 = getword(s1, buf);
		gets(buf);
		if (!strlen(buf)) blank = 1;
		l2 = getword(s2, buf);
		
		printf("%2d. ",cases ++);
		if (blank) {
			printf("Blank!\n");
			blank = 0;
		}else {
			for (int i = 0 ; i <= l1 ; ++ i)
			for (int j = 0 ; j <= l2 ; ++ j)
				f[i][j] = 0;
			for (int i = 1 ; i <= l1 ; ++ i)
			for (int j = 1 ; j <= l2 ; ++ j) {
				f[i][j] = f[i-1][j];
				if (f[i][j] < f[i][j-1])
					f[i][j] = f[i][j-1];
				if (!strcmp(s1[i-1], s2[j-1]))
					f[i][j] = f[i-1][j-1]+1;
			}	
			printf("Length of longest match: %d\n",f[l1][l2]);
		}
	}
	
	return 0;
}

UVa 10100 - Longest Match