Codeforces 628F 最大流转最小割

感觉和昨天写了的题一模一样。。。 这种题也能用hall定理取check， 感觉更最小割差不多。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e4 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, b, q, sum[N];

int cost[N][5];
int L[N], R[N];
int num[N];
int segn;

int calc(int n, int r) {
    return (n + r) / 5;
}

int main() {
    memset(sum, -1, sizeof(sum));
    scanf("%d%d%d", &n, &b, &q);
    sum[b] = n;
    sum[0] = 0;
    for(int i = 1; i <= q; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        if(~sum[x] && sum[x] != y) return puts("unfair"), 0;
        sum[x] = y;
    }
    int pre = 0;
    for(int i = 1; i <= b; i++) {
        if(~sum[i]) {
            segn++;
            L[segn] = pre + 1;
            R[segn] = i;
            num[segn] = sum[i] - sum[pre];
            if(num[segn] < 0) return puts("unfair"), 0;
            pre = i;
        }
    }
    for(int i = 1; i <= segn; i++) {
        for(int r = 0; r < 5; r++) {
            cost[i][r] = (R[i] + r) / 5 - (L[i] - 1 + r) / 5;
        }
    }
    int maxflow = inf;
    for(int mask = 0; mask < (1 << 5); mask++) {
        int ret = 0;
        for(int i = 0; i < 5; i++)
            if(mask >> i & 1) ret += n / 5;
        for(int j = 1; j <= segn; j++) {
            int tmp = 0;
            for(int i = 0; i < 5; i++)
                if(!(mask >> i & 1)) tmp += cost[j][i];
            ret += min(tmp, num[j]);
        }
        chkmin(maxflow, ret);
    }
    puts(maxflow == n ? "fair" : "unfair");
    return 0;
}

/*
*/