HDU 3879 Base Station(最大权闭合)

题目大意：给出N个站的修建花费和M条收益边，问如何建站才能使收益达到最大化

解题思路：最大权闭合，点和边分成两个点集，边和源点相连，权值为收益

点和汇点相连，权值为建造该点的花费

点和边如果有关系的话，就连边，权值为INF

具体证明的话，详见胡伯涛的最小割在信息竞赛中的应用

ISAP解法：

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 55010;
const int MAXEDGE = 400000;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
    int u, v, next;
    Type cap, flow;
    Edge() {}
    Edge(int u, int v, Type cap, Type flow, int next) {
        this->u = u;
        this->v = v;
        this->cap = cap;
        this->flow = flow;
        this->next=  next;
    }
};

struct ISAP {
    int n, m, s, t;
    Edge edges[MAXEDGE];
    int head[MAXNODE], p[MAXNODE], num[MAXNODE], cur[MAXNODE], d[MAXNODE];
    bool vis[MAXNODE];

    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof(head));
        m = 0;
    }

    void AddEdge(int u, int v, Type cap) {
        edges[m] = Edge(u, v, cap, 0, head[u]);
        head[u] = m++;
        edges[m] = Edge(v, u, 0, 0, head[v]);
        head[v] = m++;
    }

    bool BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        for (int i = 0; i < n; i++)
            d[i] = INF;
        d[t] = 0;
        vis[t] = 1;
        Q.push(t);

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

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

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

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

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

        memset(num, 0, sizeof(num));
        for (int i = 0; i < n; i++)
            cur[i] = head[i];

        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;//纪录是否找到了下一个点
            for (int i = cur[u]; ~i; i = edges[i].next) {
                Edge &e = edges[i];
                if (e.cap > e.flow && d[u] == d[e.v] + 1) {
                    ok = true;
                    p[e.v] = i;//点v由第i条边增广得到
                    cur[u] = i;//尝试到第i条边
                    u = e.v;
                    break;
                }
            }

            //如果没找到下一个点,表示u到t的最短路要变长了,或者没路可走了
            if (!ok) {
                //找寻u到下一个点的最短路
                int Min = n - 1;
                for (int i = head[u]; ~i; i = edges[i].next) {
                    Edge &e = edges[i];
                    if (e.cap > e.flow)
                        Min = min(Min, d[e.v]);
                }
                if (--num[d[u]] == 0)//GAP优化
                    break;
                num[d[u] = Min + 1]++;
                cur[u] = head[u];
                //返回前一个点，因为该点的最短距离已经变了
                if (u != s)
                    u = edges[p[u]].u;
            }
        }
        return flow;
    }
}isap;

int n, m;
void init (){
    int source = 0, sink = n + m + 1;
    int SumProfit = 0, SumCost = 0 ;

    isap.init(sink + 1);
    int u, v, c;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &c);
        isap.AddEdge(i, sink, c);
        SumCost += c;
    }

    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &u, &v, &c);
        SumProfit += c;
        isap.AddEdge(source, n + i, c);
        isap.AddEdge(n + i, u, INF);
        isap.AddEdge(n + i, v, INF);
    }
    isap.AddEdge(n + m + 2, source, SumProfit);
    isap.AddEdge(sink, n + m + 3, SumCost);
    printf("%d\n", SumProfit - isap.Maxflow(n + m + 2, n + m + 3));
}

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

Dinic解法：

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 60000;
const int MAXEDGE = 400000;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge{
    int u, v, next;
    Type cap, flow;
    Edge() {}
    Edge(int u, int v, Type cap, Type flow, int next) : u(u), v(v), cap(cap), flow(flow), next(next){}
};

struct Dinic{
    int n, m, s, t;
    Edge edges[MAXEDGE];
    int head[MAXNODE];
    int cur[MAXNODE];
    bool vis[MAXNODE];
    Type d[MAXNODE];
    vector<int> cut;

    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof(head));
        m = 0;
    }

    void AddEdge(int u, int v, Type cap) {
        edges[m] = Edge(u, v, cap, 0, head[u]);
        head[u] = m++;
        edges[m] = Edge(v, u, 0, 0, head[v]);
        head[v] = m++;
    } 

    bool BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;

        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            for (int i = head[u]; i != -1; i = edges[i].next) {
                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 = edges[i].next) {
            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] = head[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);
        }
    }
}dinic;

int n, m;

void init() {
    int source = 0, sink = n + m + 1;
    dinic.init(sink + 1);
    int Sum = 0;

    int u, v, c;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &c);
        dinic.AddEdge(m + i, sink, c);
    }

    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &u, &v, &c);
        dinic.AddEdge(source, i, c);
        dinic.AddEdge(i, v + m, INF);
        dinic.AddEdge(i, u + m, INF);
        Sum += c;
    }
    int maxflow = dinic.Maxflow(source, sink);
    printf("%d\n", Sum - maxflow);
}

int main() {
    while (scanf("%d%d", &n, &m) == 2) {
        init();
    }
    return 0;
}