題目鏈接
題意:有a,b兩台機器,有n個任務,在a機器和b機器需要不同時間,給定m個限制,如果u, v在不同機器需要額外開銷,問最小開銷
思路:最小割,源點為a機器,匯點為b機器,這樣的話求出最小割,就是把點分成兩個集合的最小代價,然後如果u, v在不同機器需要開銷,則連u,v v,u兩條邊,容量為額外開銷,這樣如果這條邊是割邊,則a,b會在不同集合,並且代價就會被加上去
代碼:
#include#include #include #include using namespace std; const int MAXNODE = 20005; const int MAXEDGE = 1000005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; } }; struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; int cur[MAXNODE]; vector cut; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; Type flow = 0; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } void MinCut() { cut.clear(); for (int i = 0; i < m; i += 2) { if (vis[edges[i].u] && !vis[edges[i].v]) cut.push_back(i); } } } gao; int n, m; int main() { while (~scanf("%d%d", &n, &m)) { gao.init(n + 2); int a, b, w; for (int i = 1; i <= n; i++) { scanf("%d%d", &a, &b); gao.add_Edge(0, i, a); gao.add_Edge(i, n + 1, b); } while (m--) { scanf("%d%d%d", &a, &b, &w); gao.add_Edge(a, b, w); gao.add_Edge(b, a, w); } printf("%d\n", gao.Maxflow(0, n + 1)); } return 0; }