Description
In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germany will fulfill the criteria, our government has so many wonderful options (raise taxes, sell stocks, revalue the gold reserves,...) that it is really hard to choose what to do.Input
The input will contain several test cases.Output
For each test case, print the longest common subsequence of words occuring in the two texts. If there is more than one such sequence, any one is acceptable. Separate the words by one blank. After the last word, output a newline character.Sample Input
die einkommen der landwirte sind fuer die abgeordneten ein buch mit sieben siegeln um dem abzuhelfen muessen dringend alle subventionsgesetze verbessert werden # die steuern auf vermoegen und einkommen sollten nach meinung der abgeordneten nachdruecklich erhoben werden dazu muessen die kontrollbefugnisse der finanzbehoerden dringend verbessert werden #
Sample Output
die einkommen der abgeordneten muessen dringend verbessert werden
題意 求兩端文本的最長公共子單詞序列 直接lcs增量法可以得出 打印路勁也是直接遞歸就行
#include#include using namespace std; const int N = 105, L = 35; char a[N][L], b[N][L]; int pre[N][N], d[N][N], n, m, la, lb, flag; void lcs() { memset (d, 0, sizeof (d)); memset (pre, 0, sizeof (pre)); for (int i = 1; i < la; ++i) for (int j = 1; j < lb; ++j) if (!strcmp (a[i], b[j])) { d[i][j] = d[i - 1][j - 1] + 1; pre[i][j] = 1; } else if (d[i - 1][j] > d[i][j - 1]) { d[i][j] = d[i - 1][j]; pre[i][j] = 2; } else { d[i][j] = d[i][j - 1]; pre[i][j] = 3; } } void print (int i, int j) { if (pre[i][j] == 1) { print (i - 1, j - 1); if (flag) flag = 0; else printf (" "); printf ("%s", a[i]); } else if (pre[i][j] == 2) print (i - 1, j); else if (pre[i][j] == 3) print (i, j - 1); } int main() { while (scanf ("%s", a[1]) != EOF) { la = 1; lb = 0; flag = 1; while (scanf ("%s", a[++la]), a[la][0] != '#'); while (scanf ("%s", b[++lb]), b[lb][0] != '#'); lcs(); print (la - 1, lb - 1); printf ("\n"); } return 0; }