题意:一共有N个城市,一些城市里有金矿,一些城市里有仓库,金矿和仓库都有一个容量,有M条边,每条边是双向的,有一个权值,求将所有金矿里的储量都运送到仓库中,所需要经过的道路中,使最大的权值最小
思路:增设一个超级源点和一个超级汇点,源点与每一个城市相连,容量为黄金数量,汇点与仓库相连,容量为仓库的容量,然后就是二分最小的最大相邻距离,跑最大流验证。
#include<stdio.h> #include<string.h> const int N=410; const int inf=0x3fffffff; int dis[N],gap[N],head[N],num,start,end,ans,n,m; int v1[N],v2[N]; struct edge { int st,ed,flow,next; }E[210000]; struct node { int x,y,w; }P[100000]; void addedge(int x,int y,int w) { E[num].st=x;E[num].ed=y;E[num].flow=w;E[num].next=head[x];head[x]=num++; E[num].st=y;E[num].ed=x;E[num].flow=0;E[num].next=head[y];head[y]=num++; } int dfs(int u,int minflow) { if(u==end)return minflow; int i,v,f,flow=0,min_dis=ans-1; for(i=head[u];i!=-1;i=E[i].next) { if(E[i].flow) { v=E[i].ed; if(dis[v]+1==dis[u]) { f=dfs(v,E[i].flow>minflow-flow?minflow-flow:E[i].flow); E[i].flow-=f; E[i^1].flow+=f; flow+=f; if(flow==minflow)break; if(dis[start]>=ans)return flow; } min_dis=min_dis>dis[v]?dis[v]:min_dis; } } if(flow==0) { if(--gap[dis[u]]==0) dis[start]=ans; dis[u]=min_dis+1; gap[dis[u]]++; } return flow; } int isap() { int maxflow=0; memset(dis,0,sizeof(dis)); memset(gap,0,sizeof(gap)); gap[0]=ans; while(dis[start]<ans) maxflow+=dfs(start,inf); return maxflow; } void makemap(int D) { memset(head,-1,sizeof(head));num=0; int i,x,y; for(i=1;i<=n;i++) { addedge(i,i+n,inf); addedge(start,i,v1[i]); addedge(i+n,end,v2[i]); } for(i=0;i<m;i++) { x=P[i].x;y=P[i].y; if(P[i].w<=D) { addedge(x+n,y,inf); addedge(y+n,x,inf); } } } int main() { int i,total,sum,flag,L,R,mid; while(scanf("%d",&n),n) { total=sum=0; for(i=1;i<=n;i++) { scanf("%d",&v1[i]); sum+=v1[i]; } for(i=1;i<=n;i++) { scanf("%d",&v2[i]); total+=v2[i]; } if(sum>total){printf("No Solution\n");continue;} scanf("%d",&m); for(i=0;i<m;i++) scanf("%d%d%d",&P[i].x,&P[i].y,&P[i].w); start=n+n+1;end=n+n+2;ans=end;L=0;R=10000;flag=-1; while(L<=R) { mid=(L+R)/2; makemap(mid); if(isap()==sum) { flag=mid; R=mid-1; } else L=mid+1; } if(flag==-1)printf("No Solution\n"); else printf("%d\n",flag); } return 0; }