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  • 网络流24题(01)搭配飞行员(最大流)

    思路:

    设置源点为 0,汇点为 n + 1,源点向飞行员引弧,容量为 1,飞行员和副飞行员引弧,容量为 1,副飞行员向 汇点引弧,容量为 1,求最大流。

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <vector>
    using namespace std;
    
    const int MAXN = 110;
    const int INFS = 0x3FFFFFFF;
    
    struct edge {
        int from, to, cap, flow;
        edge(int _from, int _to, int _cap, int _flow) 
            : from(_from), to(_to), cap(_cap), flow(_flow) {}
    };
    
    struct ISAP {
        vector<edge> edges;
        vector<int> G[MAXN];
        int s, t, n;
        int d[MAXN], p[MAXN], cur[MAXN], gap[MAXN];
        bool vis[MAXN];
    
        void clearall(int n) {
            this->n = n;
            edges.clear();
            for (int i = 0; i < n; i++)
                G[i].clear();
        }
    
        void addedge(int u, int v, int cap) {
            edges.push_back(edge(u, v, cap, 0));
            edges.push_back(edge(v, u, 0, 0));
            G[u].push_back(edges.size() - 2);
            G[v].push_back(edges.size() - 1);
        }
    
        bool BFS() {
            queue<int> Q;
            memset(d, 0, sizeof(d));
            memset(vis, false, sizeof(vis));
            vis[t] = true;
            Q.push(t);
    
            while (!Q.empty()) {
                int x = Q.front(); Q.pop();
                for (int i = 0; i < G[x].size(); i++) {
                    edge& e = edges[G[x][i]^1];
                    if (e.cap > e.flow && !vis[e.from]) {
                        vis[e.from] = true;
                        d[e.from] = d[x] + 1;
                        Q.push(e.from);
                    }
                }
            }
            return vis[s];
        }
    
        int augment() {
            int x = t, f = INFS;
            while (x != s) {
                edge& e = edges[p[x]];
                f = min(f, e.cap - e.flow);
                x = e.from;
            }
            x = t;
            while (x != s) {
                edges[p[x]].flow += f;
                edges[p[x]^1].flow -= f;
                x = edges[p[x]].from;
            }
            return f;
        }
    
        int maxflow(int s, int t) {
            this->s = s, this->t = t;
            BFS();
            memset(cur, 0, sizeof(cur));
            memset(p, 0, sizeof(p));
            memset(gap, 0, sizeof(gap));
            for (int i = 0; i < n; i++)
                gap[d[i]] += 1;
            int x = s, flow = 0;
            while (gap[s] < n) {
                if (x == t) {
                    flow += augment();
                    x = s;
                }
                bool flag = false;
                for (int i = cur[x]; i < G[x].size(); i++) {
                    edge& e = edges[G[x][i]];
                    if (d[x] == d[e.to] + 1 && e.cap > e.flow) {
                        flag = true;
                        cur[x] = i;
                        p[e.to] = G[x][i];
                        x = e.to; break;
                    }
                }
                if (!flag) {
                    int m = n - 1;
                    for (int i = 0; i < G[x].size(); i++) {
                        edge& e = edges[G[x][i]];
                        if (e.cap > e.flow) m = min(m, d[e.to]);
                    }
                    if (--gap[d[x]] == 0) break;
                    gap[d[x] = m+1] += 1;
                    cur[x] = 0;
                    if (x != s) x = edges[p[x]].from;
                }
            }
            return flow;
        }
    };
    
    ISAP sap;
    
    int main() {
        freopen("flyer.in", "r", stdin);
        freopen("flyer.out", "w", stdout);
    
        int n, n1;
        scanf("%d%d", &n, &n1);
        int u, v;
        sap.clearall(n + 2);
        while (scanf("%d%d", &u, &v) != EOF) {
            sap.addedge(u, v, 1);
        }
    
        int s = 0, t = n + 1;
        for (int i = 1; i <= n1; i++)
            sap.addedge(s, i, 1);
        for (int i = n1+1; i <= n; i++)
            sap.addedge(i, t, 1);
    
        printf("%d\n", sap.maxflow(s, t));
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/kedebug/p/3025147.html
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