动物森友会【科大讯飞杯L题】【二分答案+最大流】

题目链接

  有N个物品，一周有7天，然后呢，要对应的每个物品都要达到各自的需求数量，于是问，最少需要几天才可以达到要求？

  很明显的，这是线性关系的，我们可以用二分答案来解决这个问题，然后呢怎么知道是否满足条件也就是来确定的，要满足每个物品都要拿满。好了，这样不就有点构成最大流的性质了嘛（因为N很小），然后就是源点向N个物品链接流为的边，然后物品向它对应的每一天链接边，然后对应每一天向汇点链接它的总贡献物资数量。

  记得开long long。

#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 0x3f3f3f3f
#define Big_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)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 10, maxM = 2e4 + 7;
int N, E, S, T, head[maxN], cnt, cur[maxN];
ll need;
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; 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;
struct node
{
    int C, M, a[8];
    void In()
    {
        scanf("%d%d", &C, &M);
        for(int i=1; i<=M; i++) scanf("%d", &a[i]);
    }
} t[maxN];
inline void init()
{
    cnt = 0;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
}
inline bool check(ll lim)
{
    init();
    for(int i=1; i<=N; i++) _add(S, i, t[i].C);
    ll week = lim / 7;
    int day = (int)(lim - week * 7);
    for(int i=1; i<=7; i++)
    {
        if(i <= day) _add(N + i, T, 1LL * (week + 1) * E);
        else _add(N + i, T, 1LL * week * E);
    }
    for(int i=1; i<=N; i++)
    {
        for(int j=1; j<=t[i].M; j++)
        {
            _add(i, t[i].a[j] + N, INF);
        }
    }
    return mf.max_flow() == need;
}
int main()
{
    scanf("%d%d", &N, &E);
    S = 0; T = N + 7 + 1; mf.node = T + 1; need = 0;
    init();
    for(int i=1; i<=N; i++) { t[i].In(); need += t[i].C; }
    ll L = 1, R = 7e8, mid, ans = 0;
    while(L <= R)
    {
        mid = (L + R) >> 1LL;
        if(check(mid))
        {
            R = mid - 1;
            ans = mid;
        }
        else L = mid + 1;
    }
    printf("%lld\n", ans);
    return 0;
}