poj 3267 The Cow Lexicon

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table[i]表示以i为结尾的字符串最优解

则有 table[i] = min(table[i-1]+1, table[j] + 删除字符数) j, j+1, j+2...i删掉字符后能够和词典中的单词匹配

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <cstring>#include <stack>#include <cctype>#include <utility>#include <map>#include <string>#include <climits> #include <set>#include <string> #include <sstream>#include <utility>#include <ctime>//#pragma comment(linker, "/STACK:1024000000,1024000000") using std::priority_queue;using std::vector;using std::swap;using std::stack;using std::sort;using std::max;using std::min;using std::pair;using std::map;using std::string;using std::cin;using std::cout;using std::set;using std::queue;using std::string;using std::istringstream;using std::getline;using std::make_pair;using std::greater;const int INFI((INT_MAX-1) >> 1);char str[610][30];int len[610];char s[310];int table[610];int main(){int w, l;while(~scanf("%d%d", &w, &l)){scanf("%s", s+1);for(int i = 1; i <= w; ++i){scanf("%s", str[i]);len[i] = strlen(str[i]);}table[0] = 0;for(int i = 1; i <= l; ++i){table[i] = INFI;for(int j = 1; j <= w; ++j)if(len[j] <= i && s[i] == str[j][len[j]-1]){int p1 = i, p2 = len[j]-1;int count = 0;while(p1 >= 1 && p2 >= 0)if(s[p1] == str[j][p2]){--p1;--p2;}else{--p1;++count;}if(p2 == -1)table[i] = min(table[i], table[p1]+count);}table[i] = min(table[i], table[i-1]+1);}printf("%d\n", table[l]);}return 0;}