UVA - 11248 Frequency Hopping(網絡流＋割)

UVA - 11248 Frequency Hopping(網絡流＋割)

題目大意：有N個點，M條有向邊，每條邊都有相應的容量。現在要求你從點1運送東西到點N，運送的量為C
如果可以運送，輸出possible
如果不能運送，看能不能只增大一條弧，使得運送成功，如果可以的話，輸出所有可增大的弧
如果都不行，另外輸出

解題思路：先跑一次最大流，最大流如果大於等於C，就不用括弧了
如果不行的話，就進行括弧。
擴大哪些弧的容量呢。答案是割邊的容量，因為最小割==最大流
我們先找出所有的割邊，如何找割邊呢，如果是dinic算法的話，就找到vis[u] == ture 而vis[v] == false 且該邊容量大於0的邊，這些邊就是割邊了
接著將割邊一條一條的擴容，只需要擴大到C的容量就可以了，然後在殘余網絡上跑最大流就可以了
還有一個問題是怎麼求殘余網絡，只需要將所有邊的容量減去流量就是殘余網絡了

#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 Dinic{
    int n, m, s, t;
    vector edges;
    vector G[N];
    bool vis[N];
    int d[N], cur[N];

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
        }
        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));
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    } 

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

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = true;
                    d[e.to] = d[u] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a) {
        if (x == t || a == 0)
            return a;

        int flow = 0, f;
        for (int i = cur[x]; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i] ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0)
                    break;
            }
        }
        return flow;
    }

    int Maxflow(int s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS()) {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }

    void clear() {
        for (int i = 0; i < edges.size(); i++)
            edges[i].flow = 0;
    }

    void remmant() {
        for (int i = 0; i < edges.size(); i++)
            edges[i].cap -= edges[i].flow;
    }

    vector MinCut() {
        vector tmp;
        for (int i = 0; i < edges.size(); i++)
            if (vis[edges[i].from] && !vis[edges[i].to] && edges[i].cap > 0) tmp.push_back(i);
        return tmp;
    }
};

Dinic dinic;
#define M 10010
int n, m, c;
int cas = 1;
struct Cut{
    int u, v;
    bool operator <(const Cut &a) const {
        if (u == a.u)
            return v < a.v;
        return u < a.u;
    }
}cut[M];

int cmp(const Cut &a, const Cut &b) {
    if (a.u == b.u)
        return a.v < b.v;
    return a.u < b.u;
}

void solve() {
    dinic.init(n);
    int u, v, cost;
    for (int i = 0; i < m; i++) {
        scanf(%d%d%d, &u, &v, &cost);
        dinic.AddEdge(u, v, cost);
    }
    printf(Case %d: , cas++);
    int Maxflow = dinic.Maxflow(1, n);
    if (Maxflow >= c) printf(possible
);
    else {
        int cnt = 0;
        vector tmp = dinic.MinCut();
        dinic.remmant();
        for (int i = 0; i < tmp.size(); i++) {
            Edge &e = dinic.edges[tmp[i]];
            dinic.clear();
            e.cap = c;
            int ans = dinic.Maxflow(1,n);
            if (ans + Maxflow >= c) { 
                cut[cnt].u  = dinic.edges[tmp[i]].from;
                cut[cnt++].v = dinic.edges[tmp[i]].to;
            }
            e.cap = 0;
        }

        if (cnt == 0)
            printf(not possible
);
        else {
            sort(cut, cut + cnt);
            printf(possible option:(%d,%d), cut[0].u, cut[0].v);
            for (int i = 1; i < cnt; i++) printf(,(%d,%d), cut[i].u, cut[i].v);
            printf(
);
        }
    }
}

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