hdu 3315 My Brute 费用流,费用最小且代价最小
很常见的想法了= =
#include <stdio.h> #include <iostream> #include <string.h> using namespace std; const int N=400; const int MAXE=200000; const int inf=1<<30; int head[N],s,t,cnt,n,m,ans; int d[N],pre[N]; bool vis[N]; int q[MAXE]; int V[N],H[N],P[N],A[N],B[N]; struct Edge { int u,v,c,w,next; }edge[MAXE]; void addedge(int u,int v,int w,int c) { edge[cnt].u=u; edge[cnt].v=v; edge[cnt].w=w; edge[cnt].c=c; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].v=u; edge[cnt].u=v; edge[cnt].w=-w; edge[cnt].c=0; edge[cnt].next=head[v]; head[v]=cnt++; } int SPFA() { int l,r; memset(pre,-1,sizeof(pre)); memset(vis,0,sizeof(vis)); for(int i=0;i<=t;i++) d[i]=inf; d[s]=0; l=0;r=0; q[r++]=s; vis[s]=1; while(l<r) { int u=q[l++]; vis[u]=0; for(int j=head[u];j!=-1;j=edge[j].next) { int v=edge[j].v; if(edge[j].c>0&&d[u]+edge[j].w<d[v]) { d[v]=d[u]+edge[j].w; pre[v]=j; if(!vis[v]) { vis[v]=1; q[r++]=v; } } } } if(d[t]==inf) return 0; return 1; } void MCMF() { int flow=0; while(SPFA()) { int u=t; int mini=inf; while(u!=s) { if(edge[pre[u]].c<mini) mini=edge[pre[u]].c; u=edge[pre[u]].u; } flow+=mini; u=t; ans+=d[t]*mini; while(u!=s) { edge[pre[u]].c-=mini; edge[pre[u]^1].c+=mini; u=edge[pre[u]].u; } } } int beat(int i,int j) { int x,y; if(P[j]%A[i]==0) x=P[j]/A[i]; else x=P[j]/A[i]+1; if(H[i]%B[j]==0) y=H[i]/B[j]; else y=H[i]/B[j]+1; if(x<=y) return 1; else return 0; } int main() { int i,j; while(1) { scanf("%d",&n); if(n==0) break; s=0;t=2*n+1;cnt=0; for(i=0;i<=t;i++) head[i]=-1; for(i=1;i<=n;i++) scanf("%d",&V[i]); for(i=1;i<=n;i++) scanf("%d",&H[i]); for(i=1;i<=n;i++) scanf("%d",&P[i]); for(i=1;i<=n;i++) scanf("%d",&A[i]); for(i=1;i<=n;i++) scanf("%d",&B[i]); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(beat(i,j)) { if(i==j) addedge(i,j+n,-(V[i]*100+1),1); else addedge(i,j+n,-V[i]*100,1); } else { if(i==j) addedge(i,j+n,V[i]*100-1,1); else addedge(i,j+n,V[i]*100,1); } } for(i=1;i<=n;i++) { addedge(s,i,0,1); addedge(i+n,t,0,1); } ans=0; MCMF(); if(ans>=0) printf("Oh, I lose my dear seaco!\n"); else printf("%d %.3f%%\n",-ans/100,-1.0*100*(ans%100)/n); } return 0; }