您的位置:首页 > 其它

poj 2112 (最大流+二分)

2013-08-12 18:44 411 查看
题意:有k台挤奶机,c头奶牛,给出这k+c个实体间的距离,求出每头奶牛都到一台挤奶机去,怎么分配使奶牛走的最大距离最小。

用二分枚举最大距离,,,,

#include<stdio.h>
#include<string.h>
#define N 500
#define inf 0x3fffffff
int map

,dis
,gap
,head
,num,n,m,D,start,end,ans;
struct edge
{
int st,ed,flow,next;
}E[N*40];
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,min_dis=ans-1,flow=0;
for(i=head[u];i!=-1;i=E[i].next)
{
if(E[i].flow>0)
{
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 DD)
{
int i,j;
memset(head,-1,sizeof(head));
num=0;
for(i=1;i<=m;i++)
addedge(start,i+n,1);
for(i=1;i<=n;i++)
addedge(i,end,D);
for(i=n+1;i<=n+m;i++)
for(j=1;j<=n;j++)
{
if(map[i][j]<=DD&&map[i][j]!=0)
addedge(i,j,1);
}
}
int main()
{
int i,j,k,left,right,mid;
while(scanf("%d%d%d",&n,&m,&D)!=-1)
{
left=0;right=0;start=0;end=n+m+1;ans=end+1;
memset(map,0,sizeof(map));
for(i=1;i<=n+m;i++)
for(j=1;j<=n+m;j++)
scanf("%d",&map[i][j]);
for(k=1;k<=n+m;k++)
{
for(i=1;i<=n+m;i++)
{
if(map[i][k]==0)continue;//0表示不连通
for(j=1;j<=n+m;j++)
{
if(map[k][j]==0||i==j)continue;
if(map[i][j]==0||map[i][j]>map[i][k]+map[k][j])
map[i][j]=map[i][k]+map[k][j];
if(right<map[i][j])
right=map[i][j];
}
}
}
while(left<right)
{
mid=(left+right)/2;
makemap(mid);
int sum=isap();
if(sum==m)
right=mid;
else left=mid+1;
}
printf("%d\n",right);
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: