最大流建图。开始以为旧桥有1000座,没敢用枚举,后来看看题目发现了只是十二座。枚举桥的状态没问题。
对于隧道的容量W,可以虚拟出第三个结点表示,如u->v。增加一个点p,u->p(INF),p->v(INF),p->End(W);
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int INF=0x3f3f3f3f; const int MAXN=300;//点数的最大值 const int MAXM=5000;//边数的最大值 struct Node{ int from,to,next; int cap; }edge[MAXM]; struct Edge{ int u,v,w; }Tun[30],Acient[20],Modern[1000]; int tol; int dep[MAXN]; int head[MAXN]; int val[MAXN]; void init(){ tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w){ edge[tol].from=u; edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; head[u]=tol++; edge[tol].from=v; edge[tol].to=u; edge[tol].cap=0; edge[tol].next=head[v]; head[v]=tol++; } int BFS(int start,int end){ int que[MAXN]; int front,rear; front=rear=0; memset(dep,-1,sizeof(dep)); que[rear++]=start; dep[start]=0; while(front!=rear){ int u=que[front++]; if(front==MAXN)front=0; for(int i= head[u];i!=-1; i=edge[i].next){ int v=edge[i].to; if(edge[i].cap>0&& dep[v]==-1){ dep[v]=dep[u]+1; que[rear++]=v; if(rear>=MAXN) rear=0; if(v==end)return 1; } } } return 0; } int dinic(int start,int end){ int res=0; int top; int stack[MAXN]; int cur[MAXN]; while(BFS(start,end)){ memcpy(cur,head, sizeof(head)); int u=start; top=0; while(1){ if(u==end){ int min=INF; int loc; for(int i=0;i<top;i++) if(min>edge [stack[i]].cap){ min=edge [stack[i]].cap; loc=i; } for(int i=0;i<top;i++){ edge[stack[i]].cap-=min; edge[stack[i]^1].cap+=min; } res+=min; top=loc; u=edge[stack[top]].from; } for(int i=cur[u]; i!=-1; cur[u]=i=edge[i].next) if(edge[i].cap!=0 && dep[u]+1==dep[edge[i].to]) break; if(cur[u] !=-1){ stack [top++]= cur[u]; u=edge[cur[u]].to; } else{ if(top==0) break; dep[u]=-1; u= edge[stack [--top] ].from; } } } return res; } int main(){ int n,e; int nT,nA,nM; int u,v,w,p; while(scanf("%d%d",&n,&e)!=EOF){ nT=nA=nM=0; for(int i=1;i<=n;i++) scanf("%d",&val[i]); for(int i=0;i<e;i++){ scanf("%d%d%d%d",&u,&v,&w,&p); if(p==0){ Modern[nM].u=u;Modern[nM].v=v; Modern[nM].w=w; nM++; } else if(p<0){ Tun[nT].u=u;Tun[nT].v=v; Tun[nT].w=w; nT++; } else{ Acient[nA].u=u; Acient[nA].v=v; Acient[nA].w=w; nA++; } } int Start=0,End; End=n+nT+1; int len=(1<<nA); int cost,people,ansc=0,anspeople=0; for(int i=0;i<len;i++){ init(); cost=0; for(int k=1;k<=n;k++){ addedge(Start,k,val[k]); } for(int k=0;k<nA;k++){ if(i&(1<<k)){ addedge(Acient[k].u,Acient[k].v,INF); cost+=Acient[k].w; } else addedge(Acient[k].u,Acient[k].v,1); } for(int k=0;k<nM;k++){ addedge(Modern[k].u,Modern[k].v,INF); } for(int k=0;k<nT;k++){ addedge(Tun[k].u,n+k+1,INF); addedge(n+k+1,Tun[k].v,INF); addedge(n+k+1,End,Tun[k].w); } // system("pause"); people=dinic(Start,End); if(people>anspeople){ anspeople=people; ansc=cost; } else if(people==anspeople){ ansc=min(cost,ansc); } } if(anspeople==0){ printf("Poor Heaven Empire "); } else printf("%d %d ",anspeople,ansc); } return 0; }