bzoj 1232 [Usaco2008Nov]安慰奶牛cheer

思路：看出跟dfs的顺序有关就很好写了， 对于一棵树来说确定了起点那么访问点的顺序就是dfs序，每个点经过

其度数遍，每条边经过2边， 那么我们将边的权值×2加上两端点的权值跑最小生成树，最后加上一个最小的点的

权值最为dfs的起点。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 1e4 + 10;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

struct Edge {
    int u, v, cost;
    bool operator < (const Edge &rhs) const {
        return cost < rhs.cost;
    }
}edge[M];

int n, m, c[N], fa[N];

int getRoot(int x) {
    return x == fa[x] ? x : getRoot(fa[x]);
}

LL kruscal() {
    sort(edge + 1, edge + 1 + m);
    for(int i = 1; i <= n; i++) fa[i] = i;
    LL ans = 0, cnt = 0;
    for(int i = 1; i <= m; i++) {
        int u = edge[i].u, v = edge[i].v, cost = edge[i].cost;
        int x = getRoot(u), y = getRoot(v);

        if(x != y) {
            cnt++;
            ans += cost;
            fa[x] = y;
            if(cnt == n - 1) break;
        }
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    int mn = inf;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &c[i]);
        mn = min(mn, c[i]);
    }

    for(int i = 1; i <= m; i++) {
        int u, v, cost;
        scanf("%d%d%d", &u, &v, &cost);
        edge[i].u = u;
        edge[i].v = v;
        edge[i].cost = 2 * cost + c[u] + c[v];
    }

    LL ans = kruscal() + mn;
    printf("%lld\n", ans);
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/9128911.html