标签：char form pos out any target ace lower do it
http://codeforces.com/contest/1015/problem/B
You are given two strings ss and tt. Both strings have length nn and consist of lowercase Latin letters. The characters in the strings are numbered from 11 to nn.
You can successively perform the following move any number of times (possibly, zero):
You can‘t apply a move to the string tt. The moves are applied to the string ss one after another.
Your task is to obtain the string tt from the string ss. Find any way to do it with at most 104104 such moves.
You do not have to minimize the number of moves, just find any sequence of moves of length 104104 or less to transform ss into tt.
The first line of the input contains one integer nn (1≤n≤501≤n≤50) — the length of strings ss and tt.
The second line of the input contains the string ss consisting of nn lowercase Latin letters.
The third line of the input contains the string tt consisting of nn lowercase Latin letters.
If it is impossible to obtain the string tt using moves, print "-1".
Otherwise in the first line print one integer kk — the number of moves to transform ss to tt. Note that kk must be an integer number between 00and 104104 inclusive.
In the second line print kk integers cjcj (1≤cj<n1≤cj<n), where cjcj means that on the jj-th move you swap characters scjscj and scj+1scj+1.
If you do not need to apply any moves, print a single integer 00 in the first line and either leave the second line empty or do not print it at all.
6
abcdef
abdfec
4
3 5 4 5
4
abcd
accd
-1
代码：
#include <bits/stdc++.h>
using namespace std;
int N;
char A[55], B[55];
vector<int> T;
map<char, int> mp;
int main() {
mp.clear();
scanf("%d", &N);
scanf("%s%s", A, B);
for(int i = 0; i < N; i ++) {
mp[A[i]] ++;
mp[B[i]] --;
}
for(int i = 0; i < N; i ++) {
if(mp[A[i]]) {
printf("-1\n");
return 0;
}
}
for(int i = 0; i < N; i ++) {
for(int j = i; j < N; j ++) {
if(B[i] == A[j]) {
for(int k = j; k > i; k --) {
swap(A[k], A[k - 1]);
T.push_back(k);
}
break;
}
}
}
printf("%d\n", T.size());
for(int i = 0; i < T.size(); i ++) {
printf("%s", i != 0 ? " " : "");
printf("%d", T[i]);
}
printf("\n");
return 0;
}
CodeForces B. Obtaining the String
标签：char form pos out any target ace lower do it
原文地址：https://www.cnblogs.com/zlrrrr/p/9845809.html
