[USACO06FEB]Steady Cow Assignment G【二分+最大流】

题目链接 P2857 [USACO06FEB]Steady Cow Assignment G

  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.

数据范围：(1 <= N <= 1000) ，(1 <= B <= 20)

  看到这样的数据范围，很明显的是B很小，我们的答案也就是1～B这个区间了，我们甚至可以枚举答案，因为常数只有20，当然，我们也可以二分这个答案，这样就更小了。

  我们二分答案mid，然后枚举范围是1～mid还是2～mid+1还是3～mid+2以此类推，来判断答案的可行性，如果最大流为N那么答案就是可行的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 27, maxM = 1e5 + 7;
int N, B, mp[1005][25], cap[25];
int S, T, head[maxN], cnt, cur[maxN];
struct Eddge
{
    int nex, to; ll flow;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
    int gap[maxN], d[maxN], que[maxN], ql, qr, node;
    inline void init()
    {
        for(int i=0; i<=node + 1; i++)
        {
            gap[i] = d[i] = 0;
            cur[i] = head[i];
        }
        ++gap[d[T] = 1];
        que[ql = qr = 1] = T;
        while(ql <= qr)
        {
            int x = que[ql ++];
            for(int i=head[x], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(!d[v]) { ++gap[d[v] = d[x] + 1]; que[++qr] = v; }
            }
        }
    }
    inline ll aug(int x, ll FLOW)
    {
        if(x == T) return FLOW;
        int flow = 0;
        for(int &i=cur[x], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(d[x] == d[v] + 1)
            {
                ll tmp = aug(v, min(FLOW, edge[i].flow));
                flow += tmp; FLOW -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
                if(!FLOW) return flow;
            }
        }
        if(!(--gap[d[x]])) d[S] = node + 1;
        ++gap[++d[x]]; cur[x] = head[x];
        return flow;
    }
    inline ll max_flow()
    {
        init();
        ll ret = aug(S, INF);
        while(d[S] <= node) ret += aug(S, INF);
        return ret;
    }
} mf;
inline void init()
{
    cnt = 0; S = 0; T = N + B + 1; mf.node = T + 1;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
    for(int i=1; i<=N; i++) _add(S, i, 1);
    for(int i=1; i<=B; i++) _add(N + i, T, cap[i]);
}
inline bool solve(int lim)
{
    for(int i=1; i<=B - lim + 1; i++)
    {
        init();
        for(int u=1; u<=N; u++)
        {
            for(int j=i; j<=i+lim - 1; j++)
            {
                _add(u, N + mp[u][j], 1);
            }
        }
        if(mf.max_flow() == N) return true;
    }
    return false;
}
int main()
{
    scanf("%d%d", &N, &B);
    for(int i=1; i<=N; i++)
    {
        for(int j=1; j<=B; j++)
        {
            scanf("%d", &mp[i][j]);
        }
    }
    for(int i=1; i<=B; i++) scanf("%d", &cap[i]);
    int L = 1, R = B, mid, ans = B;
    while(L <= R)
    {
        mid = (L + R) >> 1;
        if(solve(mid))
        {
            R = mid - 1;
            ans = mid;
        }
        else L = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}