POJ3267——The Cow Lexicon

The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9558   Accepted: 4585 Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometimes they hear words that do not make any sense. For instance, Bessie once received a message that said "browndcodw". As it turns out, the intended message was "browncow" and the two letter "d"s were noise from other parts of the barnyard. The cows want you to help them decipher a received message (also containing only characters in the range 'a'..'z') of length L (2 ≤ L ≤ 300) characters that is a bit garbled. In particular, they know that the message has some extra letters, and they want you to determine the smallest number of letters that must be removed to make the message a sequence of words from the dictionary. Input Line 1: Two space-separated integers, respectively: W and L Line 2: L characters (followed by a newline, of course): the received message Lines 3..W+2: The cows' dictionary, one word per line Output Line 1: a single integer that is the smallest number of characters that need to be removed to make the message a sequence of dictionary words. Sample Input 6 10 browndcodw cow milk white black brown farmer Sample Output 2 Source USACO 2007 February Silver

题目大意：

现有一个长度为L的字符串，已知W个单词，问最少删除多少个字母之后，字符串能变为单词的组合；

思路：

DP题；

用step[ i ]表示从str[ 0...i ]中最少去掉字符的个数, remove [ j, k ] 表示str[ j ... k ] 中包含了某个单词后去掉的字符个数。

状态转移方程就为 step[ i ] = min (step[i], d[ j - 1] + remove[ j, i ] )，0 < j <=i。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int w, l;
char dir[610][310];
int len[610];
char str[310];
int step[310];

int dp()
{

step[0] = 1;
for( int i = 1;i < l;i++ )
{
step[i] = step[i-1]+1; //初始化为上一步删除的字母数+1
for( int j = 1;j <= w;j++ ) //从第一个单词开始遍历
{
int x = len[j] - 1, y = i;
if( x>y ) continue;
while( x>=0&&y>=0&&x<=y )
{
if( str[y]==dir[j][x] )
x--; //若匹配，单词的指针前移
y--; //无论匹不匹配 字符串的指针都前移
}
if( x<0 )
step[i] = min(step[i], step[y] + i - y - len[j]); //y可能为-1,因为step为全局数组，所以未操作的值都为0
}
}
return step[l-1];
}

int main()
{
cin>>w>>l;
cin>>str;
for( int i = 1;i <= w;i++ )
{
cin>>dir[i];
getchar();
len[i] = strlen(dir[i]);
}

cout<<dp()<<endl;

return 0;
}