BZOJ ~ 1312 ~ Neerc2006 Hard Life （最大密度子圖）
阿新 • • 發佈：2018-12-11
做法①
轉化為補集思想，求最小割做法。
//#include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #include<vector> using namespace std; const int MAXN = 1e5 + 5; const int INF = 0x3f3f3f3f; struct Edge { int from, to; double cap, flow; //起點,終點,容量,流量 Edge(int u, int v, double c, double 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, double 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)); memset(d, 0, sizeof(d)); 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]; } double DFS(int x, double a)//沿阻塞流增廣 { if (x == t || a == 0) return a; double 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; } double MaxFlow(int s, int t) { this->s = s; this->t = t; double flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } }solve; const int maxn = 1005; const double eps = 1e-7; int n, m, u[maxn], v[maxn], d[maxn]; bool check(double x) { double tot = 0; int s = 0, t = n+m+1; solve.init(t); for (int i = 1; i <= n; i++) solve.AddEdge(i, t, x); for (int i = 1; i <= m; i++) { solve.AddEdge(s, i+n, 1); tot += 1; solve.AddEdge(i+n, u[i], INF); solve.AddEdge(i+n, v[i], INF); } double MF = solve.MaxFlow(s, t); return tot-MF > eps;//注意這裡必須為eps不能為0 } int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= m; i++) { scanf("%d%d", &u[i], &v[i]); d[u[i]]++; d[v[i]]++; } if (m == 0) { printf("1\n"); return 0; } double l = 1.0/n, r = m; while (r-l > eps) { double mid = (l+r)/2; if (check(mid)) l = mid; else r = mid; } check(l); int ans = 0; for (int i = 1; i <= n; i++) if (solve.d[i]) ans++; printf("%d\n", ans); /*for (int i = 1; i <= n; i++) if (solve.d[i]) printf("%d\n", i);*/ return 0; } /* 5 6 1 5 5 4 4 2 2 5 1 2 3 1 */
做法②：
//#include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #include<vector> using namespace std; const int MAXN = 1e5 + 5; const int INF = 0x3f3f3f3f; struct Edge { int from, to; double cap, flow; //起點,終點,容量,流量 Edge(int u, int v, double c, double 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, double 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)); memset(d, 0, sizeof(d));//清空層次圖，因為輸出方案要用 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]; } double DFS(int x, double a)//沿阻塞流增廣 { if (x == t || a == 0) return a; double 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; } double MaxFlow(int s, int t) { this->s = s; this->t = t; double flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } }solve; const int maxn = 1005; const double eps = 1e-7; int n, m, u[maxn], v[maxn], d[maxn]; bool check(double x) { double tot = 0; int s = 0, t = n+m+1; solve.init(t); for (int i = 1; i <= n; i++) solve.AddEdge(i, t, x); for (int i = 1; i <= m; i++) { solve.AddEdge(s, i+n, 1); tot += 1; solve.AddEdge(i+n, u[i], INF); solve.AddEdge(i+n, v[i], INF); } double MF = solve.MaxFlow(s, t); return tot-MF > eps;//注意這裡必須為eps不能為0 } int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= m; i++) { scanf("%d%d", &u[i], &v[i]); d[u[i]]++; d[v[i]]++; } if (m == 0) { printf("1\n"); return 0; } double l = 1.0/n, r = m; while (r-l > eps) { double mid = (l+r)/2; if (check(mid)) l = mid; else r = mid; } check(l); int ans = 0; for (int i = 1; i <= n; i++) if (solve.d[i]) ans++; printf("%d\n", ans); /*for (int i = 1; i <= n; i++) if (solve.d[i]) printf("%d\n", i);*/ return 0; } /* 5 6 1 5 5 4 4 2 2 5 1 2 3 1 */