• [模板]洛谷P3381 最小费用最大流(dinic)


    #include<bits/stdc++.h>
    #define ll long long
    #define PB push_back
    #define endl '\n'
    #define INF 0x3f3f3f3f
    #define LINF 0x3f3f3f3f3f3f3f3f
    #define ull unsigned long long
    #define lson rt << 1, l, mid
    #define rson rt << 1 | 1, mid + 1, r
    #define lowbit(x) (x & (-x))
    #define rep(i, a, b) for(int i = a ; i <= b ; ++ i)
    #define per(i, a, b) for(int i = b ; i >= a ; -- i)
    #define clr(a, b) memset(a, b, sizeof(a))
    #define in insert
    #define random(x) (rand()%x)
    #define PII(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define pi acos(-1)
    using namespace std;
    const int maxn = 2e6 + 1000;
    const ll mod = 998244353;
    const int M = 50000 + 10;
    const int N = 5000 + 10;
    int n, m, s, t, cnt;
    int head[N], vis[N], inq[N], cur[N];
    ll dis[N];
    ll maxflow, cost;
    struct node{
        int to, next;
        ll cost, flow;
    }edge[M<<1];
    void add(int u, int v, int cost, int flow){
        edge[cnt].to = v;
        edge[cnt].cost = cost;
        edge[cnt].flow = flow;
        edge[cnt].next = head[u];
        head[u] = cnt ++;
    }
    bool spfa(){
        rep(i, 1, n) dis[i] = 1e18, inq[i] = 0, cur[i] = head[i];
        queue<int> que; dis[s] = 0;
        inq[s] = 1; que.push(s);
        while(!que.empty()){
            int u = que.front();
            que.pop();
            inq[u] = 0;
            for(int i = head[u] ; ~ i ; i = edge[i].next){
                int v = edge[i].to;
                if(dis[v] > dis[u] + edge[i].cost && edge[i].flow){
                    dis[v] = dis[u] + edge[i].cost;
                    if(!inq[v]){
                        inq[v] = 1;
                        que.push(v);
                    }
                }
            }
        }
        return dis[t] == 1e18 ? 0 : 1;
    }
    ll dfs(int u, ll flow){
        if(u == t){
            vis[t] = 1;
            maxflow += flow;
            return flow;
        }
        ll used = 0;
        vis[u] = 1;
        for(int i = cur[u] ; ~ i ; i = edge[i].next){
            int v = edge[i].to;
            cur[u] = i;
            if((v == t || !vis[v]) && edge[i].flow && dis[v] == dis[u] + edge[i].cost){
                ll minflow = dfs(v, min(flow - used, edge[i].flow));
                if(minflow){
                    cost += minflow * edge[i].cost;
                    edge[i].flow -= minflow;
                    edge[i^1].flow += minflow;
                    used += minflow;
                }
                if(used == flow) break;
            }
        }
        return used;
    }
    void dinic(){
        while(spfa()){
            vis[t] = 1;
            while(vis[t]){
                clr(vis, 0);
                dfs(s, 1e18);
            }
        }
    }
    signed main(){
        ll u, v, w, z; int q;
        scanf("%d %d %d %d", &n, &m, &s, &t);
        clr(head, -1); clr(vis, 0);
        maxflow = cost = 0;
        //s = 1; t = n;
        cnt = 0;
        while(m --){
            scanf("%lld %lld %lld %lld", &u, &v, &w, &z);
            add(u, v, z, w);
            add(v, u, -z, 0);
        }
        dinic();
        cout << maxflow << ' ' << cost << endl;
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/Ketchum/p/13294673.html
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