POJ2112 Optimal Milking（最大流）

先Floyd求牛到机器最短距离，然后二分枚举最长的边。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #include<algorithm>
  5 using namespace std;
  6 #define INF (1<<30)
  7 #define MAXN 233
  8 #define MAXM 233*233*20
  9
 10 struct Edge{
 11     int v,cap,flow,next;
 12 }edge[MAXM];
 13 int vs,vt,NE,NV;
 14 int head[MAXN];
 15
 16 void addEdge(int u,int v,int cap){
 17     edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
 18     edge[NE].next=head[u]; head[u]=NE++;
 19     edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
 20     edge[NE].next=head[v]; head[v]=NE++;
 21 }
 22
 23 int level[MAXN];
 24 int gap[MAXN];
 25 void bfs(){
 26     memset(level,-1,sizeof(level));
 27     memset(gap,0,sizeof(gap));
 28     level[vt]=0;
 29     gap[level[vt]]++;
 30     queue<int> que;
 31     que.push(vt);
 32     while(!que.empty()){
 33         int u=que.front(); que.pop();
 34         for(int i=head[u]; i!=-1; i=edge[i].next){
 35             int v=edge[i].v;
 36             if(level[v]!=-1) continue;
 37             level[v]=level[u]+1;
 38             gap[level[v]]++;
 39             que.push(v);
 40         }
 41     }
 42 }
 43
 44 int pre[MAXN];
 45 int cur[MAXN];
 46 int ISAP(){
 47     bfs();
 48     memset(pre,-1,sizeof(pre));
 49     memcpy(cur,head,sizeof(head));
 50     int u=pre[vs]=vs,flow=0,aug=INF;
 51     gap[0]=NV;
 52     while(level[vs]<NV){
 53         bool flag=false;
 54         for(int &i=cur[u]; i!=-1; i=edge[i].next){
 55             int v=edge[i].v;
 56             if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
 57                 flag=true;
 58                 pre[v]=u;
 59                 u=v;
 60                 //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
 61                 aug=min(aug,edge[i].cap-edge[i].flow);
 62                 if(v==vt){
 63                     flow+=aug;
 64                     for(u=pre[v]; v!=vs; v=u,u=pre[u]){
 65                         edge[cur[u]].flow+=aug;
 66                         edge[cur[u]^1].flow-=aug;
 67                     }
 68                     //aug=-1;
 69                     aug=INF;
 70                 }
 71                 break;
 72             }
 73         }
 74         if(flag) continue;
 75         int minlevel=NV;
 76         for(int i=head[u]; i!=-1; i=edge[i].next){
 77             int v=edge[i].v;
 78             if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
 79                 minlevel=level[v];
 80                 cur[u]=i;
 81             }
 82         }
 83         if(--gap[level[u]]==0) break;
 84         level[u]=minlevel+1;
 85         gap[level[u]]++;
 86         u=pre[u];
 87     }
 88     return flow;
 89 }
 90
 91 int d[MAXN][MAXN],k,c,m,N;
 92 bool available(int mx){
 93     memset(head,-1,sizeof(head));
 94     for(int i=1; i<=k; ++i) addEdge(i,vt,m);
 95     for(int i=1; i<=c; ++i) addEdge(vs,i+k,1);
 96     for(int i=1; i<=c; ++i){
 97         for(int j=1; j<=k; ++j){
 98             if(d[i+k][j]==0 || d[i+k][j]>mx) continue;
 99             addEdge(i+k,j,1);
100         }
101     }
102     return ISAP()==c;
103 }
104 inline void floyd(){
105     for(int k=1; k<=N; ++k){
106         for(int i=1; i<=N; ++i){
107             if(!d[i][k]) continue;
108             for(int j=1; j<=N; ++j){
109                     if(d[k][j] && (d[i][j]==0 || d[i][j]>d[i][k]+d[k][j])) d[i][j]=d[i][k]+d[k][j];
110             }
111         }
112     }
113 }
114 int main(){
115     scanf("%d%d%d",&k,&c,&m);
116     N=k+c;
117
118     vs=0; vt=N+1; NV=vt+1;
119
120     for(int i=1; i<=N; ++i){
121         for(int j=1; j<=N; ++j) scanf("%d",&d[i][j]);
122     }
123     floyd();
124
125     int l=0,r=INF;
126     while(l<r){
127         int mid=l+r>>1;
128         if(available(mid)) r=mid;
129         else l=mid+1;
130     }
131     printf("%d",l);
132     return 0;
133 }