POJ 2987 Firing 最大权闭合子图

来源：互联网发布：如何卸载2008数据库编辑：程序博客网时间：2024/06/05 15:09

参考资料：《最小割模型在信息学竞赛中的应用》

最大权闭合子图：

若一有向图G的一个子V图满足 V中所有点的出边仍指向V中的点，则该子图V就是G的一个闭合子图，其中V的顶点权和最大的称为G的最大权闭合子图

上图中闭合图有：{5}、{2,5}、 {2,4,5}、{3,4,5}、 {1,2,3,4,5}、{1,2,4,5}

最大权闭合子图为 {3,4,5}

建图方法：

新加源点S和汇点T，对于所有权值大于0的点，连一条从S到该点的边，容量为该点的权值，对于所有权值小于0的点，连一条从该点到T的边，容量为该点的权值的绝对值，对于原图中的边，建边，容量为INF，如下图所示。

最大权闭合子图的权值 = 所有正权点和 - 最大流

以割为界，靠近S一侧的点属于最大权闭合子图中的点

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <vector>using namespace std;#define REP(i, a, b) for(int i = a; i < b; i++)#define FOR(i, a, b) for(int i = a; i <= b; i++)#define CLR(a, x) memset(a, x, sizeof(a))#define bug puts("***bug***")typedef long long LL;const int maxn = 5000 + 10;const int INF = 1e9;struct Edge{          int u, v, cap, flow;          Edge(){}          Edge(int u, int v, int cap, int flow) : u(u), v(v), cap(cap), flow(flow){}};struct ISAP{          int n, s, t;          int cur[maxn], d[maxn], p[maxn], num[maxn], vis[maxn];          vector<int> G[maxn];          vector<Edge> edges;          void init(int n){                    this -> n = n;                    REP(i, 0, n) G[i].clear();                    edges.clear();                    CLR(vis, 0);          }          void add(int u, int v, int cap){                    edges.push_back(Edge(u, v, cap, 0));                    edges.push_back(Edge(v, u, 0, 0));                    int m = edges.size();                    G[u].push_back(m-2);                    G[v].push_back(m-1);          }          void bfs(){                    REP(i, 0, n) d[i] = INF;                    d[t] =0;                    queue<int> Q;                    Q.push(t);                    while(!Q.empty()){                              int x = Q.front(); Q.pop();                              REP(i, 0, G[x].size()){                                        Edge &e = edges[G[x][i]];                                        if(e.cap > 0 || d[e.v] <= d[x] + 1) continue;                                        d[e.v] = d[x] + 1;                                        Q.push(e.v);                              }                    }          }          int augment(){                    int x = t, a = INF;                    while(x != s){                              Edge &e = edges[p[x]];                              a = min(a, e.cap - e.flow);                              x = e.u;                    }                    x = t;                    while(x != s){                              edges[p[x]].flow += a;                              edges[p[x]^1].flow -= a;                              x = edges[p[x]].u;                    }                    return a;          }          LL maxflow(int s, int t){                    this -> s = s;                    this -> t = t;                    CLR(cur, 0); CLR(num, 0);                    bfs();                    REP(i, 0, n) if(d[i] != INF) num[d[i]]++;                    LL flow = 0;                    int x = s;                    while(d[s] < n){                              if(x == t){                                        flow += (LL)augment();                                        x = s;                              }                              int ok = 0;                              REP(i, cur[x], G[x].size()){                                        Edge &e = edges[G[x][i]];                                        if(e.cap > e.flow && d[e.v] + 1 == d[x]){                                                  ok = 1;                                                  cur[x] = i;                                                  p[e.v] = G[x][i];                                                  x = e.v;                                                  break;                                        }                              }                              if(!ok){                                        int m = n - 1;                                        REP(i, 0, G[x].size()){                                                  Edge &e = edges[G[x][i]];                                                  if(e.cap > e.flow) m = min(m, d[e.v]);                                        }                                        if(--num[d[x]] == 0) break;                                        ++num[d[x] = m + 1];                                        cur[x] = 0;                                        if(x != s) x = edges[p[x]].u;                              }                    }                    return flow;          }          void dfs(int x){                    vis[x] = 1;                    REP(i, 0, G[x].size()){                              Edge &e = edges[G[x][i]];                              if(e.cap > e.flow && !vis[e.v])                                        dfs(e.v);                    }          }}solver;int n, m, x, S, T, ans;LL sum, flow;void solve(){          sum = 0;          ans = 0;          solver.init(n + 2);          S = 0, T = n + 1;          FOR(i, 1, n){                    scanf("%d", &x);                    if(x > 0){                              solver.add(S, i, x);                              sum += (LL)x;                    }                    else solver.add(i, T, -x);          }          REP(i, 0, m){                    int u, v;                    scanf("%d%d", &u, &v);                    solver.add(u, v, INF);          }          flow = solver.maxflow(S, T);          solver.dfs(S);          FOR(i, 1, n) if(solver.vis[i]) ans++;          printf("%d %I64d\n", ans, sum - flow);}int main(){//          freopen("in.txt", "r", stdin);          while(~scanf("%d%d", &n, &m)){                    solve();          }          return 0;}

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