cf739E Gosha is hunting (flows)

739E

There are $ a $ ordinary balls, $ b $ super ball, there are $ n $ th to capture the treasure can dream, for the first $ i $ a treasure can dream of an ordinary ball capture probability is $ p_i $, super ball catch probability is $ u_i $, each can only throw a ball to the same treasure can dream the same dream can be a treasure can throw two kinds of balls. Then ask the number of captures in the optimal policy expectations

Consider the probability of $ dp $, we found that the state can not be reduced to $ n ^ 2 $ level,It is not dp

Assume that each treasure can only be a dream to throw a ball that matches the problem, set $ A $ as an ordinary ball, $ B $ super ball, the source point to the $ A $, $ B $ and even the capacity of the ball number, cost of $ 0 $ side, $ a, B $ respectively each sprite even a capacity of $ 1 $, cost is $ P_i $ or $ u_i $ edges and each sprite to sink even capacity of $ 1 $ at a cost of $ 0 $ sides, the biggest cost is the cost flow of the answer.

Since a wizard can simultaneously throwing two balls, when he threw two balls, the probability of capture is $ 1- (1-p_i) times (1-u_i) $, simplification was $ p_i + u_i-p_i times u_i $ , this is actually the equivalent of $ a, B $ while the capacity of $ 1 $ flows through this wizard, if there is no $ p_i times u_i $ each sprite can be connected to the sink two sides, each side capacity of $ 1 $ and the cost is $ 0 $, however, to deal with this $ p_i times u_i $ can be considered one of the two sides in the cost changed to $ -p_i times u_i $, because the maximum cost flow path of the longest running time will certainly take priority spending $ 0 $ the piece, walk $ -p_i times u_i $ of this, while walking two and only when the two balls are thrown at the same time a wizard, and this time the cost just for the $ p_i + u_i-p_i times u_i $, is proved.

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using namespace std; const double eps= 1e-8; const int maxn = 2000 + 5 ; int n,a,b; struct { int to; double cost; int cap,rev; edge( int to= 0, double cost= 0, int cap= 0, int rev= 0): to(to),cost(cost),cap(cap),rev(rev) {} }; vector<edge> g[maxn]; void addedge(int from,int to,int cap,double cost) { cost=-cost; g[from].push_back(edge(to,cost,cap,g[to].size())); g[to].push_back(edge(from,-cost, 0,g[from].size() -1)); } double p[maxn],u[maxn]; int S,T,A,B; #define MP make_pair bool inque[maxn]; double dis[maxn]; int preve[maxn],prevv[maxn]; queue< int > q; double spfa() { for( int i=S;i<=B;i++) dis[i]= 1e60,inque[i]= false; dis[S]= 0;inque[S]= true; while(!q.empty()) q.pop(); q.push(S); memset(prevv, 0, sizeof(prevv)); memset(preve, 0, sizeof(preve)); while(!q.empty()) { int u=q.front();q.pop();inque[u]= false; for( int i= 0;i<( int)g[u].size();i++){ edge &e=g[u][i]; if(e.cap && dis[e.to]-(dis[u]+e.cost)>eps) { dis[e.to]=dis[u]+e.cost; preve[e.to]=i,prevv[e.to]=u; if(!inque[e.to]){ inque[e.to]= true; q.push(e.to); } } } } if(dis[T]>= 1e60) return 0; int gap=INT_MAX; for( int i=T;i!=S;i=prevv[i]) gap=min(gap,g[prevv[i]][preve[i]].cap); for( int i=T;i!=S;i=prevv[i]){ edge &e=g[prevv[i]][preve[i]]; e.cap-=gap; g[i][e.rev].cap+=gap; } return dis[T]*( double)gap; } double MaxcostMaxflow() { double res= 0,ret= 0; while((res=spfa())!= 0) ret+=res,res= 0; return ret; } int main() { scanf( "%d%d%d",&n,&a,&b); for( int i= 1;i<=n;i++) scanf( "%lf",&p[i]); for( int i= 1;i<=n;i++) scanf( "%lf",&u[i]); S= 0,T=n+ 1,A=n+ 2,B=n+ 3; addedge(S,A,a, 0),addedge(S,B,b, 0); for( int i= 1;i<=n;i++) addedge(A,i, 1,p[i]),addedge(B,i, 1,u[i]),addedge(i,T, 1, 0),addedge(i,T, 1,-p[i]*u[i]); printf( "%.4lfn",-MaxcostMaxflow()); return 0; }

原文:大专栏  cf739E Gosha is hunting (flows)