Ultraviolet A-1368 (greedy, string simulation), and ultraviolet A-1368 greedy

Ultraviolet A-1368 (greedy, string simulation), and ultraviolet A-1368 greedy

Click Open Link

This is a string simulation question with a greedy idea. Given m strings with a length of n, you can find a string with a length of n, the number and minimum number of characters corresponding to the m string.

To minimize the number of characters in the corresponding position, each character in the string takes precedence over the characters with more occurrences at the position. If the number of occurrences is the same, the character with a smaller Lexicographic Order is selected.

Code:

# Include <iostream> # include <cstdio> # include <string. h> # include <map> # include <stack> # include <queue> # include <algorithm> # include <math. h >#include <vector >#include <set> # define from (I, a, n) for (int I = a; I <n; I ++) # define refrom (I, n, a) for (int I = n; I> = a; I --) # define EPS 1e-10 # define mod 1000000007 using namespace std; const double INF = 0x3f3f3f; const int MAX = 1000 + 10; int m, n, pos, num; char s [52] [MAX], s1 [MAX], s2 [51]; // s1 is used to record the smallest qualified sequence in the Lexicographic Order, and s2 is used to process the character int main () {int T; cin> T at each position; while (T --) {cin> m> n; int dis = 0; memset (s1, 0, sizeof (s1); num = 0; // getchar (); from (I, 0, m) cin> s [I]; from (I, 0, n) {pos = 0; int maxlen =-INF, cnt = 1; char c; from (j, 0, m) s2 [pos ++] = s [j] [I]; // extract all characters at position I from sort (s2, s2 + pos); // sort all characters in ascending order, then the first character found must be a small lexicographic character c = s2 [0]; from (j, 1, pos) {if (s2 [j] = s2 [J-1]) cnt ++; else {if (maxlen <cnt) {maxlen = cnt; c = s2 [J-1];} // changes only when the number of characters appears in hours, otherwise, the condition cnt = 1;} if (maxlen <cnt) {maxlen = cnt; c = s2 [pos-1];} // judge s1 [num ++] = c; dis ++ = (m-maxlen) after jumping out of the loop ); // The total number of characters minus the maximum number of occurrences is the contribution of the characters in the position to the total distance} cout <s1 <endl <dis <endl ;}return 0 ;}

Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.