POJ 3267 The Cow Lexicon（DP）

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题意：给出一个字符串和一个字典，求出最少删除这个字符串中的几个字符使得字符串是字典中单词的一个组合。

思路：DP。用dp[i]表示以i为后缀的字符串中最少删除几个字符使得字符串是字典中单词的一个组合，那么我们就可以得到状态转移方程dp[i] = min(dp[i], dp[pos+1]+pos-i+1-strlen(word))，初始化dp[i] = dp[i+1] + 1。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<ctime>#define eps 1e-6#define LL long long#define pii pair<int, int>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 350; int dp[MAXN], n, m, len[700];char str[700][50], T[MAXN];int main() {    //freopen("input.txt", "r", stdin);while(scanf("%d", &m) == 1) {scanf("%d", &n);cin >> T+1;for(int i = 1; i <= m; i++) {scanf("%s", str[i]+1);len[i] = strlen(str[i]+1);}dp[n+1] = 0;for(int i = n; i; i--) {dp[i] = dp[i+1] + 1; for(int j = 1; j <= m; j++) {if(str[j][1] != T[i]) continue;int pos = i;for(int k = 1; k<=len[j] && pos <= n; k++) {while(pos<=n && T[pos]!=str[j][k]) pos++;if(pos<=n && T[pos]==str[j][k]) {if(k == len[j]) dp[i] = min(dp[i], dp[pos+1]+pos-i+1-len[j]);pos++;}}}}//for(int i = 1; i <= n; i++) cout << dp[i] << endl;cout << dp[1] << endl;}    return 0;}

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