POJ 2987 Firing
题意:n个人,每个人被炒由于都会得到一个利益(可正可负),现在有一些下属关系,如果一个人被炒了,他的下属要一起炒掉,问怎么炒使得炒利益最大,炒的人最少
思路:最大权闭合的题,要处理最少,那么其实就是在求最小割分成两个集合的时候,尽量让点都到T集合去,那么只要从S进行一次dfs,遇到满流的边就停止,经过的点都给S,就是要炒掉的人了
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXNODE = 5005; const int MAXEDGE = 200005; typedef long long Type; const Type INF = 0x3f3f3f3f3f3f3f; 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<int> 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<int> 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; } int dfs(int u) { vis[u] = true; int ans = 1; for (int i = first[u]; i + 1; i = next[i]) { int v = edges[i].v; if (edges[i].flow == edges[i].cap) continue; if (vis[v]) continue; ans += dfs(v); } return ans; } void solve(Type tot) { Type ans = Maxflow(0, n - 1); memset(vis, false, sizeof(vis)); printf("%d %lld\n", dfs(0) - 1, tot - ans); } } gao; typedef long long ll; int n, m; ll sum; int main() { while (~scanf("%d%d", &n, &m)) { ll w; sum = 0; gao.init(n + 2); for (int i = 1; i <= n; i++) { scanf("%lld", &w); if (w < 0) gao.add_Edge(i, n + 1, -w); if (w > 0) { gao.add_Edge(0, i, w); sum += w; } } int u, v; while (m--) { scanf("%d%d", &u, &v); gao.add_Edge(u, v, INF); } gao.solve(sum); } return 0; }
时间: 2024-10-07 09:46:06