alice& #39; s chance

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• Serie A champions:
  Alice is going to make a film. There are n films, which stipulate that Alice I film will be able to make movies in only a few days of a week, and only in the previous WI weeks, and this film will be the end of a di days. Ask Alice if she can make a movie in full.
  The first line represents how many sets of data, for each group of data the first row represents the N movie, the next 2 to n+1 lines, each row represents a movie, 9 numbers per row, 7 numbers in front, 1 for the clap. 0 means no, the 8th number represents a few days, and the 9th number represents a couple of weeks to shoot
  
• Analysis:
  For every one of the movies. DI-X[0]-x[1]-...-x[t] = 0,di Indicates the number of days required for this movie, x[] Indicates whether the movie is selected for these days (value zero or one)
  For every single day. Y[0]-y[1]-...-y[t] <= said that it was the day to make a movie. Y[] Indicates whether the corresponding movie is taken on this day (value zero or one)

struct edge{int from, to, cap, flow; BOOL operator< (const edge& RHS) Const {return from < Rhs.from | |    (from = = Rhs.from && to < rhs.to);    }};const int MAXV = 500;struct isap{int n, m, S, t;    Vector<edge> edges;   Vector<int> G[MAXV]; adjacency table.        G[I][J] indicates that the J-side of node I in the e array ordinal bool VIS[MAXV];           BFS using int D[MAXV];        Distance from starting point to I int CUR[MAXV];          Current arc pointer int P[MAXV];        Can be augmented on the road on an arc int NUM[MAXV];        Distance designator count void Addedge (int from, int. to, int cap) {Edges.push_back (Edge) {from, to, Cap, 0});        Edges.push_back (Edge) {To, from, 0, 0});        m = Edges.size ();        G[from].push_back (m-2);    G[to].push_back (m-1);        } bool BFS () {memset (Vis, 0, sizeof (VIS));        Queue<int> Q;        Q.push (t);        Vis[t] = 1;        D[t] = 0; while (!            Q.empty ()) {int x = Q.front ();            Q.pop (); REP (i, g[x].size()) {edge& e = edges[g[x][i]^1];                    if (!vis[e.from] && e.cap > E.flow) {vis[e.from] = 1;                    D[e.from] = d[x] + 1;                Q.push (E.from);    }}} return vis[s];        } void ClearAll (int n) {this->n = n;        REP (i, N) g[i].clear ();    Edges.clear ();    } void Clearflow () {REP (I, Edges.size ()) Edges[i].flow = 0;        } int Augment () {int x = t, a = INF;            while (x! = s) {edge& E = edges[p[x]];            A = min (A, e.cap-e.flow);        x = Edges[p[x]].from;        } x = t;            while (x! = s) {Edges[p[x]].flow + = A;            Edges[p[x]^1].flow-= A;        x = Edges[p[x]].from;    } return A;        } int Maxflow (int s, int t, int need) {this->s = s;        This->t = t;int flow = 0;        BFS ();        memset (num, 0, sizeof (num));        REP (i, n) num[d[i]]++;        int x = s;        memset (cur, 0, sizeof (cur));                while (D[s] < n) {if (x = = t) {flow + = Augment ();                if (flow >= need) return flow;            x = S;            } int ok = 0;                FF (i, cur[x], g[x].size ()) {edge& E = edges[g[x][i]];                    if (E.cap > E.flow && d[x] = = D[e.to] + 1)//Advance {OK = 1;                    P[e.to] = G[x][i]; CUR[X] = i;                    Note x = e.to;                Break }} if (!ok)//Retreat {int m = n-1;//Initial value attention REP (I, g[x]                    . Size ()) {edge& E = edges[g[x][i]]; if (E.cap > E.flow) m = min (M, d[e.To]);                } if (--num[d[x]] = = 0) break;                Num[d[x] = m + 1]++; Cur[x] = 0;            Note if (x! = s) x = Edges[p[x]].from;    }} return flow;        } vector<int> Mincut ()//Call the after Maxflow {BFS ();        Vector<int> ans;            REP (i, edges.size ()) {edge& e = edges[i];        if (!vis[e.from] && vis[e.to] && e.cap > 0) ans.push_back (i);    } return ans;    } void Reduce () {REP (I, Edges.size ()) edges[i].cap-= Edges[i].flow;        } void Print () {printf ("graph:\n");    REP (i, edges.size ()) printf ("%d->%d,%d,%d\n", Edges[i].from, Edges[i].to, Edges[i].cap, Edges[i].flow);    }} mf;int Day[18];int main () {int T, n, D, W;    RI (T);        FE (Kase, 1, T) {int sum = 0;        RI (n); Mf. CLEARALL (n+ 7 * 50 + 2);        int S = 0, T = n + 7 * + 1, Max =-1;            Fe (i, 1, n) {Fe (J, 1, 7) RI (Day[j]);            RII (d, W);            sum + = D;            max = MAX (max, W); Mf.            Addedge (S, I, D); FE (J, 1, 7) if (Day[j]) {REP (k, W) MF.                Addedge (i, n + k * 7 + J, 1); }} FE (I, 1, 7) REP (J, Max) MF.        Addedge (n + J * 7 + I, T, 1); printf ("%s\n", MF. Maxflow (S, T, INF) = = sum?    "Yes": "No"); } return 0;}

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alice& #39; s chance