【题解】LuoGu3482：[POI2009]SLO-Elephants

Portal original title
first affirmed looking ring, the end of the enumeration, looking forward, because this is only one way to ensure that before
the ring to find out, take a look at how to calculate the cost of
a ring set up to find the sum of the $sum$ , the number of elements in the ring$len$ , the smallest elementin thering$Min$ , the smallest element in the entire input$minw$

• $len=1.\; The$ entire ring is just one element, pointing to yourself, and nothing is required
• $len-2.$ Two elements point to each other, just exchange it, the cost is$sum$
• $len>2.$ Simulate it on the draft paper and find that it can be understood that each element is exchanged once to its target position, and there is an element on the way as a transit to help another$len-One$ element exchange, and at the same time exchange yourself. The element is naturally the smallest, namely$Min$ , its degree, total$len-Once$ , it can also be understood that when all elements are exchanged once, it exchanges more$len-2$ times. So the cost is$sum+(len-2)Min$
  But there is another solution, we can take an element outside the ring as a transit, so the outsider must meet the minimum input, namely$minw$ , and then calculate the cost. I won’t repeat it here. It is$sum+Min+(len+1)minw$

Code：

#include <bits/stdc++.h>
#define maxn 1000010
#define LL long long
using namespace std;
LL w[maxn], ans, sum, len, Min, minw;
int a[maxn], b[maxn], pre[maxn], vis[maxn], n;

inline int read(){
    
    
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void dfs(int u){
    
    
	vis[u] = 1, sum += w[a[u]], ++len, Min = min(Min, w[a[u]]);
	if (!vis[pre[b[u]]]) dfs(pre[b[u]]);
}

int main(){
    
    
	n = read();
	minw = 1e9;
	for (int i = 1; i <= n; ++i) minw = min(minw, w[i] = read());
	for (int i = 1; i <= n; ++i) pre[a[i] = read()] = i;
	for (int i = 1; i <= n; ++i) b[i] = read();
	for (int i = 1; i <= n; ++i)
		if (!vis[i]){
    
    
			sum = 0, len = 0, Min = 1e9;
			dfs(i);
			if (len == 1);
			else if (len == 2) ans += sum;
			else ans += sum + min((len - 2) * Min, (len + 1) * minw + Min);
		}
	printf("%lld\n", ans);
	return 0;
}