【题解】LuoGu2941：[USACO09FEB]环绕岛屿Surround the Islands_p2941 [usaco09feb]环绕岛屿surround the island-CSDN博客

本文介绍了一种通过图的缩点技术来简化复杂度并求解最短路径问题的方法。具体步骤包括构建图、进行Tarjan算法缩点、计算各节点间的最短距离，并最终得出整体的最小往返距离。

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原题传送门
直接缩点，缩好后也不用建新图了，数据小
边权取最小值
对于每个新点，往外做一次，综合ans
最后ans*2，因为往返

Code：

#include <bits/stdc++.h>
#define maxn 1010
#define LL long long
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
int num, head[maxn], dfn[maxn], low[maxn], Index, vis[maxn], top, sta[maxn];
int dis[maxn][maxn], n, color[maxn], tot;

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 addedge(int x, int y){ edge[++num] = (Edge) { y, head[x] }; head[x] = num; }

void tarjan(int u){
	dfn[u] = low[u] = ++Index;
	vis[u] = 1, sta[++top] = u;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else
		if (vis[v]) low[u] = min(low[u], dfn[v]);
	}
	if (dfn[u] == low[u]){
		++tot;
		while (sta[top + 1] != u)
			color[sta[top]] = tot, vis[sta[top--]] = 0;
	}
}

int main(){
	n = read();
	for (int i = 1; i <= n; ++i){
		int x = read(), y = read();
		addedge(x, y); addedge(y, x);
	}
	for (int i = 1; i <= n; ++i)
		if (!dfn[i]) tarjan(i);
	memset(dis, 0x3f, sizeof(dis));
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j){
			int x = read();
			dis[color[i]][color[j]] = min(dis[color[i]][color[j]], x);
		}
	LL ans = 1e9;
	for (int i = 1; i <= tot; ++i){
		LL sum = 0;
		for (int j = 1; j <= tot; ++j) 
			if (i != j) sum += dis[i][j];
		ans = min(ans, sum);
	}
	printf("%lld\n", ans << 1);
	return 0;
}