POJ_3267_The Cow Lexicon_动态规划

准备给新生上课的ppt好花时间。

给一个长字符串和一个字典，问字符串中最少删除多少个字符，剩下的串可以由字典中的单词不重叠拼成。

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.

这个题不难但是我用了好几种奇怪的姿势最后看题解才做出来，一维DP，dp[i]表示从i到长串末尾最少删多少个字符满足条件，递推是这样的：

dp[i]=min( dp[i+1], dp[tem]+tem-i-tlen )

dp[i+1]是说如果舍弃i位置的字符；tem是尝试从i开始匹配一个单词，一旦匹配成功就记录tem为成功时长串的位置，tlen是匹配的字典中单词的长度。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>using namespace std;#define inf 0x3f3f3f3f#define mxl 310#define mxw 610char a[mxl];char word[mxw][30];int w,l;int dp[mxl];int main(){while(scanf("%d%d",&w,&l)!=EOF){scanf("%s",a);for(int i=0;i<w;++i)scanf("%s",word[i]);dp[l]=0;for(int i=l-1;i>=0;--i){dp[i]=dp[i+1]+1;int cur1,cur2,teml;for(int j=0;j<w;++j)if(word[j][0]==a[i]){cur1=i,cur2=0,teml=strlen(word[j]);while(cur2!=teml&&cur1!=l){if(a[cur1]==word[j][cur2])++cur2;++cur1;}if(cur2==teml)dp[i]=min(dp[i],dp[cur1]+cur1-teml-i);}}printf("%d\n",dp[0]);}return 0;}