Steady Cow Assignment

  1 #include<iostream>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cstdio>
  5 #include<queue>
  6 #define INF 100000000
  7 using namespace std;
  8 const int N=1105;
  9 const int M=1000000;
 10 struct Node{
 11    int v;
 12    int f;
 13    int next;
 14 }edge[M];
 15 int n,m,u,v,cnt,sx,ex;
 16 int head[N],pre[N];
 17 int val[N][40],val1[N];//根据题目要求申请
 18 void init(){
 19     cnt=0;
 20     memset(head,-1,sizeof(head));
 21 }
 22 void add(int u,int v,int w){
 23     edge[cnt].v=v;
 24     edge[cnt].f=w;
 25     edge[cnt].next=head[u];
 26     head[u]=cnt++;
 27     edge[cnt].f=0;
 28     edge[cnt].v=u;
 29     edge[cnt].next=head[v];
 30     head[v]=cnt++;
 31 }
 32 bool BFS(){
 33   memset(pre,0,sizeof(pre));
 34   pre[sx]=1;
 35   queue<int>Q;
 36   Q.push(sx);
 37  while(!Q.empty()){
 38      int d=Q.front();
 39      Q.pop();
 40      for(int i=head[d];i!=-1;i=edge[i].next    ){
 41         if(edge[i].f&&!pre[edge[i].v]){
 42             pre[edge[i].v]=pre[d]+1;
 43             Q.push(edge[i].v);
 44         }
 45      }
 46   }
 47  return pre[ex]>0;
 48 }
 49 int dinic(int flow,int ps){
 50     int f=flow;
 51      if(ps==ex)return f;
 52      for(int i=head[ps];i!=-1;i=edge[i].next){
 53         if(edge[i].f&&pre[edge[i].v]==pre[ps]+1){
 54             int a=edge[i].f;
 55             int t=dinic(min(a,flow),edge[i].v);
 56               edge[i].f-=t;
 57               edge[i^1].f+=t;
 58              flow-=t;
 59              if(flow<=0)break;
 60         }
 61 
 62      }
 63       if(f-flow<=0)pre[ps]=-1;
 64       return f-flow;
 65 }
 66 int solve(){
 67     int sum=0;
 68     while(BFS())
 69         sum+=dinic(INF,sx);
 70     return sum;
 71 }
 72 int main(){
 73     int ans,temp;
 74     while(~scanf("%d%d",&n,&m)) {
 75          init();
 76          ans=INF;sx=0,ex=n+m+1;
 77        for(int i=1;i<=n;i++)
 78         for(int j=1;j<=m;j++){
 79             scanf("%d",&temp);
 80             val[i][temp]=j;
 81         }
 82         for(int i=1;i<=m;i++)
 83           scanf("%d",&val1[i]);
 84     for(int i=1;i<=m;i++){
 85         int l=i,r=m;
 86         while(l<=r){
 87             int mid=(l+r)/2;
 88             if(mid-i>=ans){
 89                 r=mid-1;
 90                 continue;
 91             }
 92              init();
 93             for(int k=1;k<=n;k++)
 94                 add(0,k,1);
 95             for(int k=1;k<=n;k++)
 96             for(int h=1;h<=m;h++)
 97             if(val[k][h]<=mid&&val[k][h]>=i)
 98               add(k,n+h,1);
 99             for(int k=1;k<=m;k++)
100                 add(k+n,n+m+1,val1[k]);
101             if(solve()==n){
102                ans=min(ans,mid-i);
103                 r=mid-1;
104             }
105             else
106                 l=mid+1;
107          }
108       }
109       printf("%d\n",ans+1);
110     }
111       return 0;
112 }