Va 1016-Silly Sort (Replacement + greedy)

Va 1016-Silly Sort (Replacement + greedy)

Link: Ultraviolet A 1016-Silly Sort

Given a sequence with a length of n, each operation can exchange the positions of any two numbers. The cost is the sum of the two numbers, and the minimum cost is obtained to arrange the sequence in an ordered order.

Solution: The given sequence is mapped to a replacement based on the number of values to break down the replacement cycle. For each cycle, the minimum value is the minimum one-by-one exchange cost, but it should be considered, you can exchange the minimum value with the minimum value in the sequence, replace it with the minimum value, and finally change it back. Take the best of the two cases.

#include 
  
   #include 
   
    #include using namespace std;const int maxn = 1005;int n, arr[maxn], rec[maxn], pos[maxn];int rep, v[maxn];void init () {    memset(v, 0, sizeof(v));    memset(rec, -1, sizeof(rec));    rep = maxn;    for (int i = 0; i < n; i++) {        scanf("%d", &arr[i]);        pos[i] = arr[i];        rep = min(arr[i], rep);    }    sort(pos, pos + n);    for (int i = 0; i < n; i++)        rec[pos[i]] = i;    for (int i = 0; i < n; i++)        pos[i] = rec[arr[i]];    /*    for (int i = 0; i < n; i++)        printf("%d ", pos[i]);    printf("\n");    */}int solve () {    int ret = 0;    for (int i = 0; i < n; i++) {        if (v[i])            continue;        int j = i, c= 0;        int ans = 0, tmp = maxn;        while (v[j] == 0) {            tmp = min(tmp, arr[j]);            ans += arr[j];            v[j] = 1;            c++;            j = pos[j];        }        ans -= tmp;        ret += ans + min(tmp * (c-1), tmp * 2 + rep * (c+1));    }    return ret;}int main () {    int cas = 1;    while (scanf("%d", &n) == 1 && n) {        init();        printf("Case %d: %d\n\n", cas++, solve());    }    return 0;}