Description

https://loj.ac/problem/6010

Solution

第一问：每个数字拆成两个点，流量为1，边的流量也为1
第二问：点的流量为+∞ + ∞
第三问：边的流量也为+∞ + ∞

费用流即可。

Code

/************************************************
 * Au: Hany01
 * Date: Jul 12th, 2018
 * Prob: LOJ6010 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 200 << 1, maxm = 200 << 3;

int a[25][45], cost, beg[maxn], nex[maxm], v[maxm], f[maxm], w[maxm], vis[maxn], dis[maxn], id[25][45], S, T, cnt, n, m, e = 1;

inline void add(int uu, int vv, int ff, int ww, int fl = 1) {
    v[++ e] = vv, f[e] = ff, w[e] = ww, nex[e] = beg[uu], beg[uu] = e;
    if (fl) add(vv, uu, 0, -ww, 0);
}

inline bool BFS()
{
    static queue<int> q;
    For(i, 1, T) dis[i] = INF;
    Set(vis, 0), dis[S] = 0, q.push(S);
    while (!q.empty()) {
        int u = q.front(); q.pop(), vis[u] = 0;
        for (register int i = beg[u]; i; i = nex[i]) {
            if (f[i] && chkmin(dis[v[i]], dis[u] + w[i]))
                if (!vis[v[i]]) vis[v[i]] = 1, q.push(v[i]);
        }
    }
    return dis[T] != INF;
}

int DFS(int u, int flow)
{
    if (u == T) return flow;
    int res = flow, tmp;
    vis[u] = 1;
    for (register int i = beg[u]; i; i = nex[i])
        if (f[i] && !vis[v[i]] && dis[v[i]] == dis[u] + w[i]) {
            tmp = DFS(v[i], min(res, f[i]));
            f[i] -= tmp, f[i ^ 1] += tmp, cost += tmp * (LL)w[i];
            if (!(res -= tmp)) return flow;
        }
    return flow - res;
}

inline int MCMF() {
    for (cost = 0; BFS(); )
        do Set(vis, 0), DFS(S, INF); while (vis[T]);
    return cost;
}

int main()
{
#ifdef hany01
    File("loj6010");
#endif

    m = read(), n = read();
    For(i, 1, n) For(j, 1, m + i - 1) a[i][j] = read(), id[i][j] = ++ cnt;
    T = (S = (cnt << 1) + 1) + 1;

    For(i, 1, m) add(S, id[1][i], 1, 0);
    For(i, 1, n) For(j, 1, m + i - 1) {
        add(id[i][j], id[i][j] + cnt, 1, -a[i][j]);
        if (i != n) add(id[i][j] + cnt, id[i + 1][j], 1, 0), add(id[i][j] + cnt, id[i + 1][j + 1], 1, 0);
        else add(id[i][j] + cnt, T, 1, 0);
    }
    printf("%d\n", -MCMF());

    e = 1, Set(beg, 0);
    For(i, 1, m) add(S, id[1][i], 1, 0);
    For(i, 1, n) For(j, 1, m + i - 1) {
        add(id[i][j], id[i][j] + cnt, INF, -a[i][j]);
        if (i != n) add(id[i][j] + cnt, id[i + 1][j], 1, 0), add(id[i][j] + cnt, id[i + 1][j + 1], 1, 0);
        else add(id[i][j] + cnt, T, INF, 0);
    }
    printf("%d\n", -MCMF());

    e = 1, Set(beg, 0);
    For(i, 1, m) add(S, id[1][i], 1, 0);
    For(i, 1, n) For(j, 1, m + i - 1) {
        add(id[i][j], id[i][j] + cnt, INF, -a[i][j]);
        if (i != n) add(id[i][j] + cnt, id[i + 1][j], INF, 0), add(id[i][j] + cnt, id[i + 1][j + 1], INF, 0);
        else add(id[i][j] + cnt, T, INF, 0);
    }
    printf("%d\n", -MCMF());

    return 0;
}
//三五年时三五月，可怜杯酒不曾消。
//    -- 黄景仁《绮怀》