Codeforces739E. Gosha is hunting
$n \leq 2000$个东西要搬走，两个人分别能搬$a$和$b$件东西，给两个人搬走每种东西的概率，问最优决策下期望搬走多少东西。
一个人搬贡献：$p_i$，另一个人搬贡献：$q_i$，两人一起搬：$p_i+q_i-p_iq_i$。
$s$->两个人：流量分别$a,b$，费用0；两个人->每件东西：费用$p_i$或$q_i$，流量1；每件东西->$t$：两条边，一条流量1费用0，一条流量1费用$-p_iq_i$。跑最大费用流。
1 //#include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 //#include<time.h> 5 //#include<complex> 6 //#include<set> 7 #include<queue> 8 #include<algorithm> 9 #include<stdlib.h> 10 using namespace std; 11 12 #define LL long long 13 int qread() 14 { 15 char c; int s=0,f=1; while ((c=getchar())<'0' || c>'9') (c=='-') && (f=-1); 16 do s=s*10+c-'0'; while ((c=getchar())>='0' && c<='9'); return s*f; 17 } 18 19 //Pay attention to '-' , LL and double of qread!!!! 20 21 int n,a,b; 22 #define maxn 6011 23 #define maxm 50011 24 25 struct Edge{int to,next,cap,flow; double cost;}; 26 struct Network 27 { 28 Edge edge[maxm<<1]; int first[maxn],le,n; 29 void clear(int m) {le=2; n=m;} 30 void in(int x,int y,int cap,double cost) 31 {Edge &e=edge[le]; e.to=y; e.cap=cap; e.flow=0; e.cost=cost; e.next=first[x]; first[x]=le++;} 32 void insert(int x,int y,int cap,double cost) {in(x,y,cap,cost); in(y,x,0,-cost);} 33 double dis[maxn]; int que[maxn],head,tail,s,t,cur[maxn]; bool vis[maxn]; 34 bool spfa() 35 { 36 for (int i=1;i<=n;i++) dis[i]=1e18; dis[s]=0; 37 vis[s]=1; head=tail=0; que[tail++]=s; 38 while (head!=tail) 39 { 40 int x=que[head++]; if (head==maxn) head=0; 41 vis[x]=0; 42 for (int i=first[x];i;i=edge[i].next) 43 { 44 Edge &e=edge[i]; if (e.cap==e.flow) continue; 45 if (dis[e.to]-(dis[x]+e.cost)>1e-9) 46 { 47 dis[e.to]=dis[x]+e.cost; 48 if (!vis[e.to]) 49 { 50 que[tail++]=e.to; 51 if (tail==maxn) tail=0; 52 vis[e.to]=1; 53 } 54 } 55 } 56 } 57 return dis[t]<1e17; 58 } 59 double cost; 60 int dfs(int x,int a) 61 { 62 if (x==t || !a) {cost+=a*dis[t]; return a;} 63 int flow=0,f; 64 vis[x]=1; 65 for (int &i=cur[x];i;i=edge[i].next) 66 { 67 Edge &e=edge[i]; 68 if (!vis[e.to] && fabs(dis[e.to]-(dis[x]+e.cost))<1e-9 && (f=dfs(e.to,min(e.cap-e.flow,a)))>0) 69 { 70 e.flow+=f; flow+=f; 71 edge[i^1].flow-=f; a-=f; 72 if (!a) break; 73 } 74 } 75 vis[x]=0; 76 return flow; 77 } 78 double Dinic(int S,int T) 79 { 80 s=S; t=T; 81 cost=0; 82 while (spfa()) 83 { 84 for (int i=1;i<=n;i++) cur[i]=first[i]; 85 dfs(s,0x3f3f3f3f); 86 } 87 return cost; 88 } 89 // void test() 90 // { 91 // for (int i=1;i<=n;i++) 92 // for (int j=first[i];j;j=edge[j].next) 93 // { 94 // Edge &e=edge[j]; 95 // cout<<i<<"->"<<e.to<<": cap"<<e.cap<<" flow"<<e.flow<<" cost"<<e.cost<<endl; 96 // } 97 // cout<<endl; 98 // } 99 }g; 100 101 double p[maxn]; 102 int main() 103 { 104 n=qread(); a=qread(); b=qread(); 105 int A=n+1,B=A+1,s=B+1,t=s+1; g.clear(t); 106 g.insert(s,A,a,0); g.insert(s,B,b,0); 107 108 for (int i=1;i<=n;i++) scanf("%lf",&p[i]); 109 double x; 110 for (int i=1;i<=n;i++) 111 { 112 scanf("%lf",&x); 113 g.insert(A,i,1,-p[i]); g.insert(B,i,1,-x); 114 g.insert(i,t,1,0); g.insert(i,t,1,p[i]*x); 115 } 116 // g.test(); 117 printf("%.6f\n",-g.Dinic(s,t)); 118 // g.test(); 119 return 0; 120 }
