Ultraviolet A 1016-silly sort (replacement decomposition + greedy)

Ultraviolet A 1016-silly sort

Question Link

Given a sequence, numbers are different. Two numbers can be exchanged each time. The price of the exchange is the sum of the two numbers, and the minimum price of the sequence to increase is required.

Idea: using the replacement decomposition principle, we can consider each cycle of the sequence separately. For each cycle, constant exchange is sure that each number will change at least once, and then use greedy ideas, if we replace the minimum value in a loop each time, the minimum value will be-1 time in length, and the remaining number will be each time. Note that there is another possible optimization method here, it is to change the minimum value in the entire sequence to the cycle, and then change it after replacement. Both of them can be considered.

Code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;int n, a[N], b[N], id[N], Min;int main() {    int cas = 0;    while (~scanf("%d", &n) && n) {Min = 1000;for (int i = 1; i <= n; i++) {    scanf("%d", &a[i]);    Min = min(Min, a[i]);    b[i] = a[i];}sort(b + 1, b + n + 1);for (int i = 1; i <= n; i++)    id[b[i]] = i;int ans = 0;for (int i = 1; i <= n; i++) {    if (a[i]) {int cnt = 0;int sum = a[i];int now = id[a[i]];int tmp = a[i];a[i] = 0;while (a[now]) {    cnt++;    sum += a[now];    tmp = min(tmp, a[now]);    int save = now;    now = id[a[now]];    a[save] = 0;}ans += min(sum + tmp * cnt - tmp, sum + 2 * (tmp + Min) + Min * cnt - tmp);    }}printf("Case %d: %d\n\n", ++cas, ans);    }    return 0;}