POJ3189 Steady Cow Assignment(最大流)

题目大概说,有n头牛和b块草地,每头牛心中分别对每块草地都有排名,草地在牛中排名越高牛安排在那的幸福度就越小(。。。),每块草地都能容纳一定数量的牛。现在要给这n头牛分配草地,牛中的幸福度最大与幸福度最小的差值越小越好,问最小能多小。

显然又是枚举结果跑最大流看是否合法。不过,枚举幸福度的差值是做不了的,应该要枚举的是幸福度的最大值和幸福度的最小值。然后建图没啥好说的。。最后的结果要加1,因为题目说“including the endpoints”,虽然不知道什么意思。。

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #include<algorithm>
  5 using namespace std;
  6 #define INF (1<<30)
  7 #define MAXN 1111
  8 #define MAXM 44444
  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 n,b,happy[1111][22],cap[22];
 92 bool isok(int mm,int mx){
 93     vs=0; vt=n+b+1; NV=vt+1; NE=0;
 94     memset(head,-1,sizeof(head));
 95     for(int i=1; i<=n; ++i) addEdge(vs,i,1);
 96     for(int i=1; i<=n; ++i){
 97         for(int j=1; j<=b; ++j){
 98             if(mm<=happy[i][j] && happy[i][j]<=mx) addEdge(i,j+n,1);
 99         }
100     }
101     for(int i=1; i<=b; ++i) addEdge(i+n,vt,cap[i]);
102     return ISAP()==n;
103 }
104 int main(){
105     int a;
106     scanf("%d%d",&n,&b);
107     for(int i=1; i<=n; ++i){
108         for(int j=1; j<=b; ++j){
109             scanf("%d",&a);
110             happy[i][a]=j;    
111         }
112     }
113     for(int i=1; i<=b; ++i) scanf("%d",cap+i);
114     int res=INF;
115     for(int i=1; i<=b; ++i){
116         for(int j=i; j<=b; ++j){
117             if(isok(i,j)) res=min(res,j-i);
118         }
119     }
120     printf("%d",res+1);
121     return 0;
122 }

 

posted @ 2016-03-15 23:26  WABoss  阅读(303)  评论(0编辑  收藏  举报