題目傳送門

題目大意

給出一個\(n\)個點\(n\)條邊的圖，每個點有且僅有一個出邊，改變每條邊都會有對應的花費。求最小的花費使得整個圖強連通。

思路

很顯然，最後的圖就是一個環。那我們要求的答案實際上就是鏈的最大權值之和。

我們再次將問題轉換，發現就是每個點只保留一條邊，而保留的邊就是連向它的邊權最大的邊。但是我們發現這實際上還有問題，因為這樣仍可以構成一個環，那我們就選環上一個點斷開再選一條不在環上且邊權最大的邊即可。

\(\texttt{Code}\)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define ll long long
#define MAXN 100005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

ll ans;
int n,a[MAXN],c[MAXN],cm[MAXN],ncm[MAXN],vis[MAXN];

signed main(){
	read (n);
	for (Int i = 1;i <= n;++ i) read (a[i],c[i]),ans += c[i];
	for (Int i = 1;i <= n;++ i) if (!vis[i]){
		int x = i;for (;!vis[x];x = a[x]) vis[x] = i;
		if (vis[x] == i){int siz = 0;for (;~vis[x];x = a[x]) siz ++,vis[x] = -1;if (siz == n) return puts ("0"),0;}
	}
	for (Int i = 1;i <= n;++ i){
		cm[a[i]] = max (cm[a[i]],c[i]);
		if (~vis[i]) ncm[a[i]] = max (ncm[a[i]],c[i]);
	}
	for (Int i = 1;i <= n;++ i) ans -= cm[i];
	for (Int i = 1;i <= n;++ i) if (vis[i] == -1){
		int minn = 2e9;
		for (Int x = i;vis[x] == -1;x = a[x]) minn = min (minn,cm[x] - ncm[x]),vis[x] = 0;
		ans += minn;
	}
	write (ans),putchar ('\n');
	return 0;
}