HDU 6141

思路来自 FXXL

最小树形图模板用kuangbin的

/*
HDU 6141 - I am your Father!  [ 最小树形图 ]  |  2017 Multi-University Training Contest 8
题意：
	N个点M条边求最大树形图，还问权值最大的图中第N个点的父亲编号最小能是多少
	N <= 1e3, M <= 1e4
分析：
	为了使第n个点的父亲编号最小，修改权值，将所有权值扩大1000倍
	连向第N个节点的边再加上 n-其父亲的编号 的权值
	这样答案就是  ans/1000 和 n-ans%1000
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL INF = 1e18;
const int N = 1e3+5;
const int M = 1e4+5;
struct Edge {
    int u, v;
    LL cost;
} edge[M];
int pre[N], id[N], visit[N];
LL in[N];
int zhuliu(int root, int n, int m, Edge edge[])
{
    LL res = 0; int u, v;
    while (1)
    {
        for (int i = 0; i < n; i++) in[i] = INF;
        for (int i = 0; i < m; i++)
            if (edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v])
            {
                pre[edge[i].v] = edge[i].u;
                in[edge[i].v] = edge[i].cost;
            }
        for (int i = 0; i < n; i++)
            if (i != root && in[i] == INF)
                return -1;
        int tn = 0;
        memset(id, -1, sizeof(id));
        memset(visit, -1, sizeof(visit));
        in[root] = 0;
        for (int i = 0; i < n; i++)
        {
            res += in[i];
            v = i;
            while (visit[v] != i && id[v] == -1 && v != root)
            {
                visit[v] = i;
                v = pre[v];
            }
            if (v != root && id[v] == -1)
            {
                for (int u = pre[v]; u != v; u = pre[u]) id[u] = tn;
                id[v] = tn++;
            }
        }
        if (tn == 0) break;
        for (int i = 0; i < n; i++)
            if (id[i] == -1) id[i] = tn++;
        for (int i = 0; i < m;)
        {
            v = edge[i].v;
            edge[i].u = id[edge[i].u];
            edge[i].v = id[edge[i].v];
            if (edge[i].u != edge[i].v)
                edge[i++].cost -= in[v];
            else
                swap(edge[i], edge[--m]);
        }
        n = tn;
        root = id[root];
    }
    return res;
}
int t, n, m;
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < m; i++)
        {
            scanf("%d%d%lld", &edge[i].u, &edge[i].v, &edge[i].cost);
            edge[i].cost *= 1000;
            if (edge[i].v == n) edge[i].cost += n-edge[i].u;
            edge[i].cost *= -1;
            edge[i].u--;
            edge[i].v--;
        }
        LL ans = zhuliu(0, n, m, edge);
        ans = -ans;
        int fa = n - ans % 1000;
        ans /= 1000;
        printf("%lld %d
", ans, fa);
    }
}