Solution -「多校联训」数学考试

(mathcal{Description})

  Link.

  给定 (n) 个函数，第 (i) 个有 (f_i(x)=a_ix^3+b_ix^2+cx_i+d~(xin[l_i,r_i]capmathbb Z))，还有 (m) 条形如 (x_ule x_v+d) 的限制，请最大化 (sum_{i=1}^nf_i(x_i)) 或声明无解。

  (n,|l_i|,|r_i|le 100)。

(mathcal{Solution})

  很久没遇到了，压根儿没往网络流方面想 qwq。

  对于每个 (f_i)，拉一条代表 (f_i(l_i..r_i)) 的链，边权就是某个 (f) 的值的相反数；限制条件方便转化为最小割，之后直接跑最小割即可。

  (mathcal O(operatorname{Dinic}(sum(r-l),msum(r-l))))。

(mathcal{Code})

/*~Rainybunny~*/

#include <queue>
#include <cstdio>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )

#define int long long

inline int imin( const int a, const int b ) { return a < b ? a : b; }

const int MAXN = 100, IINF = 1ll << 50, BASE = 7e6;
int n, m, lid[MAXN + 5];

struct Function {
    int a, b, c, d, l, r;
    inline void read() {
        scanf( "%lld %lld %lld %lld %lld %lld" , &a, &b, &c, &d, &l, &r );
    }
    inline int operator () ( const int x ) const {
        return d + x * ( c + x * ( b + x * a ) );
    } 
} fc[MAXN + 5];

struct FlowGraph {
    static const int MAXND = 2e4 + 10, MAXEG = 2e5;
    int ecnt, bound, S, T, head[MAXND];
    struct Edge { int to, flw, nxt; } graph[MAXEG * 2];
    int ds[MAXND], curh[MAXND];

    FlowGraph(): ecnt( 1 ) {}

    inline void clear() {
        ecnt = 1;
        rep ( i, 0, bound ) head[i] = 0;
    }

    inline void operator () ( const int s, const int t, const int f ) {
        graph[++ecnt] = { t, f, head[s] }, head[s] = ecnt;
        graph[++ecnt] = { s, 0, head[t] }, head[t] = ecnt;
    }

    inline bool bfs() {
        static std::queue<int> que;
        rep ( i, 0, bound ) ds[i] = IINF;
        que.push( S ), ds[S] = 0;
        while ( !que.empty() ) {
            int u = que.front(); que.pop();
            for ( int i = head[u], v; i; i = graph[i].nxt ) {
                if ( graph[i].flw && ds[u] + 1 < ds[v = graph[i].to] ) {
                    ds[v] = ds[u] + 1, que.push( v );
                }
            }
        }
        return ds[T] != IINF;
    }

    inline int dfs( const int u, int iflw ) {
        if ( u == T ) return iflw;
        int oflw = 0;
        for ( int& i = curh[u], v; i; i = graph[i].nxt ) {
            if ( graph[i].flw && ds[u] + 1 == ds[v = graph[i].to] ) {
                int tmp = dfs( v, imin( iflw - oflw, graph[i].flw ) );
                oflw += tmp, graph[i].flw -= tmp, graph[i ^ 1].flw += tmp;
                if ( iflw == oflw ) break;
            }
        }
        if ( !oflw ) ds[u] = IINF;
        return oflw;
    }

    inline int calc( const int s, const int t ) {
        int ret = 0; S = s, T = t;
        while ( bfs() ) {
            rep ( i, 0, bound ) curh[i] = head[i];
            ret += dfs( S, IINF );
        }
        return ret;
    }
} G;

signed main() {
    freopen( "sleep.in", "r", stdin );
    freopen( "sleep.out", "w", stdout );

    int Q; scanf( "%lld", &Q );
    while ( Q-- ) {
        scanf( "%lld %lld", &n, &m ), G.clear();
        int S = 0, T = 1, node = 1;
        rep ( i, 1, n ) {
            fc[i].read();
            G( S, lid[i] = ++node, IINF );
            rep ( j, fc[i].l, fc[i].r ) {
                G( node, node + 1, BASE - fc[i]( j ) ), ++node;
            }
            G( node, T, IINF );
        }
        rep ( i, 1, m ) {
            int u, v, d; scanf( "%lld %lld %lld", &u, &v, &d );
            rep ( x, fc[u].l, fc[u].r ) {
                if ( fc[v].l <= x - d && x - d <= fc[v].r ) {
                    G( lid[u] + x - fc[u].l, lid[v] + x - d - fc[v].l, IINF );
                } else if ( x - d > fc[v].r ) {
                    G( lid[u] + x - fc[u].l, T, IINF );
                }
            }
        }
        G.bound = node;
        int ans = -G.calc( S, T );
        if ( ans <= -IINF ) puts( "mei ji ge" );
        else printf( "%lld
", ans + n * BASE );
    }
    return 0;
}