题意:每次交换两数,花费为两数和。最后得到递增序列。求最小花费。
思路:找到所有置换群。两种方法,和置换群里最小的交换或和整个数列最小的交换。
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define mem(a, n) memset(a, n, sizeof(a))
#define ALL(v) v.begin(), v.end()
#define si(a) scanf("%d", &a)
#define sii(a, b) scanf("%d%d", &a, &b)
#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pb push_back
#define eps 1e-8
const int inf = 0x3f3f3f3f, N = 1e4 + 5, MOD = 1e9 + 7;
int T, cas = 0;
int n, m, k;
bool vis[N];
struct node {
int num, id;
} a[N];
bool cmp(node a, node b) { return a.num < b.num; }
int main(){
#ifdef LOCAL
freopen("/Users/apple/input.txt", "r", stdin);
// freopen("/Users/apple/out.txt", "w", stdout);
#endif
while(si(n) != EOF) {
for(int i = 1; i <= n; i ++) si(a[i].num), a[i].id = i;
sort(a + 1, a + 1 + n, cmp);
k = a[1].num;
mem(vis, 0);
int ans = 0;
for(int i = 1; i <= n; i ++) {
int cnt = 0, idx = i, mi = inf, sum = 0;
while(vis[idx] == 0) {
vis[idx] = 1;
cnt ++;
mi = min(mi, a[idx].num);
sum += a[idx].num;
idx = a[idx].id;
}
if(sum) ans += min(sum + (cnt - 2) * mi, sum + mi + (cnt + 1) * k);
}
printf("%d\n", ans);
}
return 0;
}