CodeForces Gym 100989B LCS (B)

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给一个算LCS时得到的DP矩阵，然后根据这个矩阵让你算出两个符合条件的字符串

因为字符串长度<=25，然后可以先假设A字符串就是abcdef....，然后根据A算B就行，看B中哪个跟A一样，但是A中可能有两个或以上的字符跟B的某个是一样的，然后这时就要把这两个变成同一个，但是这个涉及到很多，因为可能还跟B中别的字符一样，所以就用并查集维护他们，把他们并到一个集合里，最后输出时按集合的根输出就行。

或者不用并查集，每次遇到这个问题时，就把AB字符串都扫一遍，遇到要改的改掉就行，反正长度 25，随便搞

#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <cstring>#include <vector>using namespace std;#define ll long longint R, C;char A[30], B[30];int DP[30][30];int set[30];int stk[30];int findset(int x){if (x == set[x])return x;elsereturn set[x] = findset(set[x]);}void unionset(int x, int y){int fx = findset(x);int fy = findset(y);if (fx == fy)return;elseset[fy] = fx;findset(y);}int main(){//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);scanf("%d%d", &R, &C);for (int i = 0; i <= R; ++i){for (int j = 0; j <= C; ++j){scanf("%d", &DP[i][j]);}}for (int i = 1; i <= R; ++i){A[i] = 'a' + i - 1;set[i] = i;}int cnt = 0;for (int j = 1; j <= C; ++j){cnt = 0;for (int i = 1; i <= R; ++i){if (DP[i][j] > DP[i - 1][j] && DP[i][j] > DP[i][j - 1]){stk[cnt++] = i;}}for (int i = 1; i < cnt; ++i)unionset(stk[0], stk[i]);}for (int i = 1; i <= R; ++i){findset(i);}for (int i = 1; i <= R; ++i)A[i] = A[findset(i)];for (int i = 1; i <= C; ++i){if (B[i] == 0)B[i] = 'z';}for (int j = 1; j <= C; ++j){cnt = 0;for (int i = 1; i <= R; ++i){if (DP[i][j] > DP[i - 1][j] && DP[i][j] > DP[i][j - 1]){B[j] = A[set[i]];}}}printf("%s\n", A + 1);printf("%s\n", B + 1);//system("pause");//while (1);return 0;}

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