UVALive ~ 3231 ~ Fair Share (公平分配問題)
阿新 • • 發佈:2018-12-10
訓練指南上的例題。
2018-8月後LA的資料好像有問題了,理論AC一波?
#include<bits/stdc++.h> using namespace std; const int MAXN = 1e5 + 5; const int INF = 0x3f3f3f3f; struct Edge { int from, to, cap, flow; //起點,終點,容量,流量 Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {} }; struct Dinic { int n, m, s, t; //結點數,邊數(包括反向弧),源點s,匯點t vector<Edge> edges; //邊表。edges[e]和edges[e^1]互為反向弧 vector<int> G[MAXN]; //鄰接表,G[i][j]表示結點i的第j條邊在edges陣列中的序號 int d[MAXN]; //從起點到i的距離(層數差) int cur[MAXN]; //當前弧下標 bool vis[MAXN]; //BFS分層使用 void init(int n) { this->n = n; edges.clear(); for (int i = 0; i <= n; i++) G[i].clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } bool BFS()//構造分層網路 { memset(vis, 0, sizeof(vis)); queue<int> Q; d[s] = 0; vis[s] = true; Q.push(s); while (!Q.empty()) { int x = Q.front(); Q.pop(); for (int i = 0; i < G[x].size(); i++) { Edge& e = edges[G[x][i]]; if (!vis[e.to] && e.cap > e.flow) { vis[e.to] = true; d[e.to] = d[x] + 1; Q.push(e.to); } } } return vis[t]; } int DFS(int x, int a)//沿阻塞流增廣 { if (x == t || a == 0) return a; int flow = 0, f; for (int& i = cur[x]; i < G[x].size(); i++)//從上次考慮的弧 { Edge& e = edges[G[x][i]]; if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)//多路增廣 { e.flow += f; edges[G[x][i]^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } int MaxFlow(int s, int t) { this->s = s; this->t = t; int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } }solve; int n, m, a[MAXN][2]; bool check(int x) { int s = 0, t = n+m+1; solve.init(t); for (int i = 1; i <= m; i++) { solve.AddEdge(s, i, 1); solve.AddEdge(i, a[i][0]+m, 1); solve.AddEdge(i, a[i][1]+m, 1); } for (int i = 1; i <= n; i++) solve.AddEdge(i+m, t, x); int ans = solve.MaxFlow(s, t); return ans == m; } int main() { int T; scanf("%d", &T); while (T--) { scanf("%d%d", &n, &m); for (int i = 1; i <= m; i++) scanf("%d%d", &a[i][0], &a[i][1]); int l = 0, r = m; while (l < r) { int mid = (l+r)/2; if (check(mid)) r = mid; else l = mid+1; } printf("%d\n", l); } return 0; } /* 3 5 6 1 2 2 3 3 4 4 5 5 1 1 3 3 2 3 2 1 2 6 6 1 2 3 4 4 6 6 5 5 3 6 3 */