P2857 [USACO06FEB]稳定奶牛分配Steady Cow Assignment

题目描述

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.

有N头牛，B个牛棚.告诉你每头牛心里牛棚的座次，即哪个牛棚他最喜欢，哪个第2喜欢， 哪个第3喜欢，等等.但牛棚容量一定，所以每头牛分配到的牛棚在该牛心中的座次有高有低.现 在求一种最公平的方法分配牛到牛棚，使所有牛中，所居牛棚的座次最高与最低的跨度最小.

输入输出格式

输入格式：

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.

输出格式：

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

输入输出样例

输入样例#1： 

6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2

输出样例#1： 

2

说明

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.

 

Solution：

　　画风诡异的题～（洛谷标签什么鬼啊！明显的最大流嘛～而且为什么死活$WA$一个点啊～）`～`

　　我的思路其实蛮简单的，直接枚举等级差和初始等级，然后在这个范围内建图，建图方法就比较常规了：源点连容量为$1$的边到每个牛，每头牛向我们枚举的等级范围中的牛舍连容量为$1$的边，每个牛舍向汇点连容量为牛舍体积的边。

　　每次跑最大流，当最大流等于牛数时，此时的等级差就是答案了。

　　（话说应该没问题吧，我都想特判了，求解！～知道的请$@$我～感激不尽）

　　（上面全是废话，感谢大鸡哥，网络流每次建图，$cnt$应该赋值为$1$，啊啊啊 !（`～`）!）

代码（非$AC$）：

 

#include<bits/stdc++.h>
#define il inline
#define debug printf("%s %d\n",__FUNCTION__,__LINE__)
#define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
using namespace std;
const int N=500005,inf=5201314;
int n,m,ans,tot,s,t=1314,mp[1005][30],val[25],to[N],net[N],w[N],h[N],cnt=1,dis[5000];

il int gi(){
    int a=0;char x=getchar();
    while(x<'0'||x>'9')x=getchar();
    while(x>='0'&&x<='9')a=(a<<3)+(a<<1)+x-48,x=getchar();
    return a;
}

il void add(int x,int y,int z){
    to[++cnt]=y,net[cnt]=h[x],h[x]=cnt,w[cnt]=z;
    to[++cnt]=x,net[cnt]=h[y],h[y]=cnt,w[cnt]=0;
}

il bool bfs(){
    queue<int>q;
    memset(dis,-1,sizeof(dis));
    q.push(s);dis[s]=0;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=h[u];i;i=net[i])
            if(dis[to[i]]==-1&&w[i]>0)dis[to[i]]=dis[u]+1,q.push(to[i]);
    }
    //For(i,1,n+m)cout<<dis[i]<<' ';cout<<endl;//debug;
    return dis[t]!=-1;
}

il int dfs(int u,int op){
    if(u==t)return op;
    int flow=0,used=0;
    for(int i=h[u];i;i=net[i]){
        int v=to[i];
        if(dis[v]==dis[u]+1&&w[i]>0){
            used=dfs(v,min(op,w[i]));
            if(!used)continue;
            flow+=used,op-=used;
            w[i]-=used,w[i^1]+=used;
            if(!op)break;
        }
    }
    if(!flow)dis[u]=-1;
    return flow;
}

il bool work(int pos,int len){
    tot=0;cnt=1;  //万恶之源
    memset(h,0,sizeof(h));
    For(i,1,n) add(s,i,1);
    For(i,1,m) add(i+n,t,val[i]);
    For(i,1,n) For(j,pos,pos+len-1) add(i,mp[i][j],1);
    while(bfs())tot+=dfs(0,inf);
    //cout<<tot<<endl;
    if(tot==n)return 1;
    return 0;
}

int main(){
    n=gi(),m=gi();
    For(i,1,n) add(s,i,1);
    For(i,1,n) For(j,1,m) mp[i][j]=gi()+n;
    For(i,1,m) val[i]=gi();
    For(i,1,m) For(j,1,m-i+1)if(work(j,i)){cout<<i;return 0;}
}