Question: There are n points (0~n-1) and m directed edges. Ask which point is the starting point to make the minimum spanning tree weight the smallest. If it can form the output weight and vertex number, otherwise output impossible.
Problem solution: minimum tree diagram + virtual root
Fortunately, I did this problem, the board has a problem.
Since which one is required as the root weight and the smallest value, enumeration will definitely not work, consider adding points, that is, virtual roots.
For this virtual root, we connect it to each point, and the weight is the sum of the weights of all the original edges +1, so as to avoid the appearance of two or more edges connected to the virtual root.
After running Zhu Liu, the smallest tree diagram came out, how to find the starting point? Starting from the virtual root, the number -m of the newly added edge is the number of the point.
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 40010;
struct Edge {
int u, v;
int cost;
}edge[MAXM];
int pre[MAXN], id[MAXN], vis[MAXN];
int in[MAXN], pos;
int zhuliu(int rt, int n, int m) {
int u, v;
int res = 0;
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;
if (edge[i].u == rt) pos = i; //起点
}
for (int i = 0; i < n; i++)
if (i != rt && in[i] == INF) return -1;//不存在最小树形图
int tn = 0;
memset(id, -1, sizeof(id));
memset(vis, -1, sizeof(vis));
in[rt] = 0;
for (int i = 0; i < n; i++) {
res += in[i];
v = i;
while (vis[v] != i && id[v] == -1 && v != rt) {
vis[v] = i;
v = pre[v];
}
if (v != rt && id[v] == -1) {
for (int u = pre[v]; u != v; u = pre[u]) id[u] = tn;
id[v] = tn++;
}
}
if (tn == 0) break;//没有有向环
for (int i = 0; i < n; i++)
if (id[i] == -1) id[i] = tn++;
for (int i = 0; i < m; i++) {
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;
rt = id[rt];
}
return res;
}
int n, m;
int main() {
while (~scanf("%d%d", &n, &m)) {
int L = 0, sum = 0;
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &edge[L].u, &edge[L].v, &edge[L].cost);
sum += edge[L++].cost;
}
sum++;
for (int i = 0; i < n; i++) {
edge[L].u = n;
edge[L].v = i;
edge[L++].cost = sum;
}
int ans = zhuliu(n, n + 1, L);
if (ans == -1 || ans >= 2 * sum) puts("impossible");
else printf("%d %d\n", ans - sum, pos - m);
puts("");
}
return 0;
}