hdu 3228 (最大流+二分)

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;}