poj 3189（枚举加最大流）

Steady Cow Assignment

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3611 Accepted: 1245

Description

Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy. 

FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn. 

Each cow gives FJ the order in which she prefers the barns. A cow's happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn's capacity is exceeded and the size of the range (i.e., one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.

Input

Line 1: Two space-separated integers, N and B 

Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i's top-choice barn, the second integer on that line is the number of the i'th cow's second-choice barn, and so on. 

Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.

Output

Line 1: One integer, the size of the minumum range of barn rankings the cows give their assigned barns, including the endpoints.

Sample Input

6 41 2 3 42 3 1 44 2 3 13 1 2 41 3 4 21 4 2 32 1 3 2

Sample Output

2

Hint

Explanation of the sample: 

Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.

题意是这样的： 农夫约翰有n个牛，m个牛棚，每个牛都有对m个牛棚的一个排序，表示自己住在其中的舒适程度。。。你的任务就是找出一个牛棚居住的安排。。。让舒适度最小的和最大的差值最小。。。。

这个题刚开始做的时候。。。构不出来图。。。没去想暴力枚举。。。问了一下别人。。然后发现大部分都是枚举做的。。然后构图都是在枚举的时候满足枚举条件的连线就刚好了。。。好吧。。。这个题不是自己真正意义上做出来的。。。。囧

#include <iostream>#include<cstdio>#include<string>using namespace std;#define inf 1<<30#define maxM 100005#define maxN 1050#define cc(m,v) memset(m,v,sizeof(m))struct node{    int u,v,f,next;}edge[maxM];int head[maxN],p,lev[maxN],cur[maxN];int que[maxM],maze[1005][30],cap[30];void ainit(){    p=0,cc(head,-1);}bool bfs(int s,int t){    int u,i,v,qin=0,qout=0;    cc(lev,0),lev[s]=1,que[qin++]=s;    while(qout!=qin){        u=que[qout++];        for(i=head[u];i!=-1;i=edge[i].next)            if(edge[i].f>0 && lev[v=edge[i].v]==0){                lev[v]=lev[u]+1,que[qin++]=v;                if(v==t) return 1;            }    }    return lev[t];}int dinic(int s,int t){    int i,k,u,f;    int flow=0,qin;    while(bfs(s,t)){        memcpy(cur,head,sizeof(head));        u=s,qin=0;        while(1){            if(u==t){                for(k=0,f=inf;k<qin;k++)                    if(edge[que[k]].f<f)                        f=edge[que[i=k]].f;                for(k=0;k<qin;k++)                    edge[que[k]].f-=f,edge[que[k]^1].f+=f;                flow+=f,u=edge[que[qin=i]].u;            }            for(i=cur[u];cur[u]!=-1;i=cur[u]=edge[cur[u]].next)                if(edge[i].f>0 && lev[u]+1==lev[edge[i].v]) break;            if(cur[u]!=-1)                que[qin++]=cur[u],u=edge[cur[u]].v;            else{                if(qin==0) break;                lev[u]=-1,u=edge[que[--qin]].u;            }        }    }    return flow;}void addedge(int u,int v,int f){    edge[p].u=u,edge[p].v=v,edge[p].f=f,edge[p].next=head[u],head[u]=p++;    edge[p].u=v,edge[p].v=u,edge[p].f=0,edge[p].next=head[v],head[v]=p++;}void cal(int n,int m){    int i,j,u,v;    for(i=1;i<=m;i++)        for(j=1;j+i-1<=m;j++){            ainit();            for(u=1;u<=n;u++) addedge(0,u,1);            for(u=1;u<=m;u++) addedge(u+n,n+m+1,cap[u]);            for(u=1;u<=n;u++)                for(v=j;v<j+i;v++)                    addedge(u,maze[u][v]+n,1);            if(dinic(0,n+m+1)==n){                printf("%d\n",i);                return ;            }        }}int main(){    int n,m,i,j;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)        for(j=1;j<=m;j++)            scanf("%d",&maze[i][j]);    for(i=1;i<=m;i++) scanf("%d",&cap[i]);    cal(n,m);    return 0;}