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网络流24题之 1738: 最小路径覆盖问题

时间:2018-08-01 22:32:56      阅读:183      评论:0      收藏:0      [点我收藏+]

标签:show   for   max   break   覆盖   space   mem   first   node   

网络流24题之 1738: 最小路径覆盖问题

最小路径覆盖问题

模板题,求一个图的最小路径覆盖,输出边数和,路径。不会输出路径的跑dinic然后把图输出来就懂了。

#include <bits/stdc++.h>

using namespace std;

int k;

struct Dinic {

    static const int MAXN = 30005 + 7;
    static const int MAXM = 1e7 + 7;
    static const int INF = 0x3f3f3f3f;

    int n, m, s, t;

    int first[MAXN], cur[MAXN], dist[MAXN], sign;

    struct Node {
        int to, flow, next;
    } edge[MAXM];

    inline void init(int start, int vertex, int ss, int tt) {
        n = vertex, s = ss, t = tt;
        for(int i = start; i <= n; i++ ) {
            first[i] = -1;
        }
        sign = 0;
    }

    inline void addEdge(int u, int v, int flow) {
        edge[sign].to = v, edge[sign].flow = flow, edge[sign].next = first[u];
        first[u] = sign++;
    }

    inline void add_edge(int u, int v, int flow) {
        addEdge(u, v, flow);
        addEdge(v, u, 0);
    }

    int match[MAXN] = {0}, vis[MAXN] = {0};

    void show_graph() {
        int cnt = 0;
        for(int i = 1; i <= k; i++ ) {
            for(int j = first[i]; ~j; j = edge[j].next) {
                int to = edge[j].to, w = edge[j].flow;
                if(to != s && edge[j ^ 1].flow) {
                    match[i] = to - k;
                }
            }
        }
        for(int i = 1; i <= k; i++ ) {
            int now = i;
            if(vis[i]) {
                continue;
            }
            while(1) {
                printf("%d", now);
                vis[now] = 1;
                now = match[now];
                if(now == 0) {
                    cnt++;
                    puts("");
                    break;
                } else {
                    printf(" ");
                }
            }
        }
        printf("%d\n", cnt);
    }

    inline int dinic() {
        int max_flow = 0;
        while(bfs(s, t)) {
            for(int i = 0; i <= n; i++ ) {
                cur[i] = first[i];
            }
            max_flow += dfs(s, INF);
        }
        return max_flow;
    }

    bool bfs(int s, int t) {
        memset(dist, -1, sizeof(dist));
        queue<int>que;
        que.push(s), dist[s] = 0;
        while(!que.empty()) {
            int now = que.front();
            que.pop();
            if(now == t) {
                return 1;
            }
            for(int i = first[now]; ~i; i = edge[i].next) {
                int to = edge[i].to, flow = edge[i].flow;
                if(dist[to] == -1 && flow > 0) {
                    dist[to] = dist[now] + 1;
                    que.push(to);
                }
            }
        }
        return 0;
    }

    int dfs(int now, int max_flow) {
        if(now == t) {
            return max_flow;
        }
        int ans = 0, next_flow = 0;
        for(int &i = cur[now]; ~i; i = edge[i].next) {
            int to = edge[i].to, flow = edge[i].flow;
            if(dist[to] == dist[now] + 1 && flow > 0) {
                next_flow = dfs(to, min(max_flow - ans, flow));
                ans += next_flow;
                edge[i].flow -= next_flow;
                edge[i ^ 1].flow += next_flow;
                if(ans == max_flow) {
                    return max_flow;
                }

            }
        }
        if(ans == 0) {
            return dist[now] = 0;
        }
        return ans;
    }

} cwl;

int main() {
    int n, m;
    while(~scanf("%d %d", &n, &m)) {
        cwl.init(0, 2 * n + 1, 0, 2 * n + 1);
        for(int i = 1; i <= n; i++ ) {
            cwl.add_edge(0, i, 1);
            cwl.add_edge(i + n, 2 * n + 1, 1);
        }
        for(int i = 1; i <= m; i++ ) {
            int u, v;
            scanf("%d %d", &u, &v);
            cwl.add_edge(u, v + n, 1);
        }
        cwl.dinic();
        k = n;
        cwl.show_graph();
    }
    return 0;
}

网络流24题之 1738: 最小路径覆盖问题

标签:show   for   max   break   覆盖   space   mem   first   node   

原文地址:https://www.cnblogs.com/Q1143316492/p/9403632.html

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