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  • luogu P2754 [CTSC1999]家园

    本题是分层图最大流问题,相当于按时间拆点,每个当前点向下一点的下一时间层连点,每一层有n+1个点

    #include<bits/stdc++.h>
    using namespace std;
    #define lowbit(x) ((x)&(-x))
    typedef long long LL;
    
    const int maxm = 5e4+5;
    const int INF = 0x3f3f3f3f;
    
    struct edge{
        int u, v, cap, flow, nex;
    } edges[maxm];
    
    int head[maxm], cur[maxm], cnt, level[6005];
    
    void init() {
        memset(head, -1, sizeof(head));
    }
    
    void add(int u, int v, int cap) {
        edges[cnt] = edge{u, v, cap, 0, head[u]};
        head[u] = cnt++;
    }
    
    void addedge(int u, int v, int cap) {
        add(u, v, cap), add(v, u, 0);
    }
    
    void bfs(int s) {
        memset(level, -1, sizeof(level));
        queue<int> q;
        level[s] = 0;
        q.push(s);
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            for(int i = head[u]; i != -1; i = edges[i].nex) {
                edge& now = edges[i];
                if(now.cap > now.flow && level[now.v] < 0) {
                    level[now.v] = level[u] + 1;
                    q.push(now.v);
                }
            }
        }
    }
    
    int dfs(int u, int t, int f) {
        if(u == t) return f;
        for(int& i = cur[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            if(now.cap > now.flow && level[u] < level[now.v]) {
                int d = dfs(now.v, t, min(f, now.cap - now.flow));
                if(d > 0) {
                    now.flow += d;
                    edges[i^1].flow -= d;
                    return d;
                }
    
            }
        }
        return 0;
    }
    
    int dinic(int s, int t) {
        int maxflow = 0;
        for(;;) {
            bfs(s);
            if(level[t] < 0) break;
            memcpy(cur, head, sizeof(head));
            int f;
            while((f = dfs(s, t, INF)) > 0)
                maxflow += f;
        }
        return maxflow;
    }
    
    int H[25], r[25], station[25][25];
    
    void run_case() {
        init();
        int n, m, k;
        cin >> n >> m >> k;
        for(int i = 1; i <= m; ++i) {
            cin >> H[i] >> r[i];
            for(int j = 0; j < r[i]; ++j) {
                cin >> station[i][j];
                if(station[i][j] == -1) station[i][j] = n+1;
            }
        }
        int s = 0, t = n+1, maxflow = 0;
        for(int times = 1; times <= 500; ++times) {
            for(int i = 0; i <= n; ++i) addedge(i+(times-1)*(n+2), i+(times)*(n+2), INF);
            addedge(t+times*(n+2), t+(times-1)*(n+2), INF);
            for(int i = 1; i <= m; ++i) 
                addedge((times-1)*(n+2)+station[i][(times-1)%r[i]], times*(n+2)+station[i][times%r[i]], H[i]);
            maxflow += dinic(s, t);
            if(maxflow >= k) {
                cout << times;
                return;
            }
        }
        cout << 0;
    }
    
    int main() {
        ios::sync_with_stdio(false), cin.tie(0);
        run_case();
        cout.flush();
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/GRedComeT/p/12305321.html
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