Meaning of the questions: they try to focus on graph cuts right to the minimum set of average side.
"Application of minimum cut model in information science contest in the" Paper of the original title.
Fractional Programming. Consideration on each side is taken as 1.
#include <bits/stdc++.h> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } const int N = 1e5 + 10; const double EPS = 1e-5; struct Edge { int v, next; double f; } edge[N]; int n, m, head[N], cnt, level[N], iter[N], x[N], y[N]; double z[N]; bool vis[N]; inline void add(int u, int v, double f) { edge[cnt].v = v; edge[cnt].f = f; edge[cnt].next = head[u]; head[u] = cnt++; edge[cnt].v = u; edge[cnt].f = f; edge[cnt].next = head[v]; head[v] = cnt++; } bool bfs() { for (int i = 1; i <= n; i++) iter[i] = head[i], level[i] = -1, vis[i] = 0; queue<int> que; que.push(1); level[1] = 0; while (!que.empty()) { int u = que.front(); que.pop(); for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].v; double f = edge[i].f; if (level[v] < 0 && f) { level[v] = level[u] + 1; que.push(v); } } } return level[n] != -1; } double dfs(int u, double f) { if (u == n) return f; double flow = 0; for (int i = iter[u]; ~i; i = edge[i].next) { iter[u] = i; int v = edge[i].v; if (level[v] == level[u] + 1 && edge[i].f) { double w = dfs(v, min(f, edge[i].f)); //if (fabs(w) < EPS) continue; f -= w; flow += w; edge[i].f -= w, edge[i^1].f += w; if (f <= 0) break; } } return flow; } double dinic() { double ans = 0; while (bfs()) ans += dfs(1, 1e8); return ans; } bool check(double mid) { memset(head, -1, sizeof(head)); cnt = 0; double flow = 0; for (int i = 1; i <= m; i++) { if (z[i] > mid) add(x[i], y[i], z[i] - mid); else flow += z[i] - mid; } flow += dinic(); return flow >= EPS; } void get_ans(int u) { vis[u] = 1; for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].v; if (!vis[v] && edge[i].f > EPS) { get_ans(v); } } } int ans[N], tol; int main() { int kase = 1; while (~scanf("%d%d", &n, &m)) { if (kase > 1) puts(""); kase = 2; for (int i = 1; i <= m; i++) { x[i] = read();y[i] = read(); int a = read();z[i] = (double)a; } double l = 0, r = 1e7; while (r - l > EPS) { double mid = (l + r) / 2.0; if (check(mid)) l = mid; else r = mid; } check(r); get_ans(1); tol = 0; for (int i = 1; i <= m; i++) { if (vis[x[i]] + vis[y[i]] == 1 || z[i] <= r) { ans[++tol] = i; } } printf("%d\n", tol); for (int i = 1; i <= tol; i++) { if (i - 1) putchar(' '); printf("%d", ans[i]); } puts(""); } return 0; }