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 }