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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU - 4289 Control (最小割 MCMF)

HDU - 4289 Control (最小割 MCMF)

編輯:C++入門知識

HDU - 4289 Control (最小割 MCMF)


題目大意:有一個間諜要將一些機密文件送到目的地
現在給出間諜的初始位置和要去的目的地,要求你在間諜的必經路上將其攔獲且費用最小

解題思路:最小割最大流的應用,具體可以看網絡流–最小割最大流

建圖的話
超級源點–起始城市,容量為INF
城市拆成兩點(u, v),容量為監視該城市的代價
能連通的城市連接,容量為INF
目的地和超級匯點相連,容量為INF

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge {
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}
};

struct ISAP {
    int p[N], num[N], cur[N], d[N];
    int t, s, n, m;
    bool vis[N];

    vector G[N];
    vector edges;

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
            d[i] = INF;
        }
        edges.clear();
    }

    void AddEdge(int from, int to, int 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() {
        memset(vis, 0, sizeof(vis));

        queue Q;
        d[t] = 0;
        vis[t] = 1;
        Q.push(t);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i] ^ 1];
                if (!vis[e.from] && e.cap > e.flow) {
                    vis[e.from] = true;
                    d[e.from] = d[u] + 1;
                    Q.push(e.from);
                }
            }
        }
        return vis[s];
    }

    int Augment() {
        int u = t, flow = INF;
        while (u != s) {
            Edge &e = edges[p[u]];
            flow = min(flow, e.cap - e.flow);
            u = edges[p[u]].from;
        }

        u = t;
        while (u != s) {
            edges[p[u]].flow += flow;
            edges[p[u] ^ 1].flow -= flow;
            u = edges[p[u]].from;
        }
        return flow;
    }

    int Maxflow(int s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        BFS();
        if (d[s] > n)
            return 0;

        memset(num, 0, sizeof(num));
        memset(cur, 0, sizeof(cur));
        for (int i = 0; i < n; i++)
            if (d[i] < INF)
                num[d[i]]++;
        int u = s;

        while (d[s] <= n) {
            if (u == t) {
                flow += Augment();
                u = s;
            }
            bool ok = false;
            for (int i = cur[u]; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[u] == d[e.to] + 1) {
                    ok = true;
                    p[e.to] = G[u][i]; 
                    cur[u] = i;
                    u = e.to;
                    break;
                }
            }

            if (!ok) {
                int Min = n;
                for (int i = 0; i < G[u].size(); i++) {
                    Edge &e = edges[G[u][i]];
                    if (e.cap > e.flow)
                        Min = min(Min, d[e.to]);
                }
                if (--num[d[u]] == 0)
                    break;
                num[d[u] = Min + 1]++;
                cur[u] = 0;
                if (u != s)
                    u = edges[p[u]].from;
            }
        }
        return flow;
    }
};

ISAP isap;
int n, m;
void solve() {
    int source = 0, sink = 2 * n + 1;
    int s, t;
    isap.init(sink);
    scanf(%d%d, &s, &t);
    isap.AddEdge(source, s, INF);
    isap.AddEdge(t + n, sink, INF);

    int x, y;
    for (int i = 1; i <= n; i++) {
        scanf(%d, &x);
        isap.AddEdge(i, i + n, x);
    }

    for (int j = 0; j < m; j++) {
        scanf(%d%d, &x, &y);
        isap.AddEdge(x + n, y, INF);
        isap.AddEdge(y + n, x, INF);
    }
    printf(%d
, isap.Maxflow(source, sink));

}

int main() {
    while (scanf(%d%d, &n, &m) != EOF) {
        solve();
    }
    return 0;
}

 

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