BJWC2018 简要题解

T1 Kakuro

有界费用流，或者模拟 +inf 的普通费用流。

#include 
#include 
#include 
#include 
#include 
#include 
#define LOG(FMT...)  // fprintf(stderr, FMT)
using namespace std;
typedef long long ll;
struct Num {
    ll d, infs;
    Num(ll d = 0, ll infs = 0) : d(d), infs(infs) {}
    bool operator<(const Num& x) const {
        if (infs != x.infs)
            return infs < x.infs;
        return d < x.d;
    }
    Num operator+(const Num& x) const { return Num(d + x.d, infs + x.infs); }
    Num operator*(ll x) const { return Num(d * x, infs * x); }
    friend Num operator*(ll x, const Num& y) { return y * x; }
    Num& operator+=(const Num& x) {
        d += x.d;
        infs += x.infs;
        return *this;
    }
};
struct Edge {
    int v, w;
    Num c;
    Edge *next, *rev;
};
const int N = 32, V = N * N, T = V - 1, S = V - 2;
const Num INF = Num(0, 1);
int n, m, nCnt;
int typ[N][N];
int a[N][N][2];
int fromRow[N][N];
Num cst[N][N][2];
bool inq[V];
Edge* pth[V];
Num dist[V];
Edge* g[V];
Edge* addEdge(int u, int v, int w, const Num& c);
void link(int u, int v, int w, const Num& c);
Num mcf();
bool spth();
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) scanf("%d", &typ[i][j]);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            if ((typ[i][j] & 1) || typ[i][j] == 4)
                scanf("%d", &a[i][j][0]);
            if (typ[i][j] & 2)
                scanf("%d", &a[i][j][1]);
        }
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            int x;
            if ((typ[i][j] & 1) || typ[i][j] == 4) {
                scanf("%d", &x);
                if (x == -1)
                    cst[i][j][0] = INF;
                else
                    cst[i][j][0] = x;
            }
            if (typ[i][j] & 2) {
                scanf("%d", &x);
                if (x == -1)
                    cst[i][j][1] = INF;
                else
                    cst[i][j][1] = x;
            }
        }
    Num curCost = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            if (typ[i][j] & 2) {
                ++nCnt;
                int cnt = 0, orij = j;
                for (++j; typ[i][j] == 4; ++j) {
                    fromRow[i][j] = nCnt;
                    ++cnt;
                }
                int used = abs(a[i][orij][1] - cnt);
                curCost += used * cst[i][orij][1];
                if (a[i][orij][1] > cnt)
                    link(S, nCnt, used, cst[i][orij][1] * -1);
                link(S, nCnt, numeric_limits<int>::max(), cst[i][orij][1]);
                --j;
            }
        }
    for (int j = 1; j <= m; ++j)
        for (int i = 1; i <= n; ++i)
            if (typ[i][j] & 1) {
                ++nCnt;
                int cnt = 0, orii = i;
                for (++i; typ[i][j] == 4; ++i) {
                    int used = abs(a[i][j][0] - 1);
                    curCost += used * cst[i][j][0];
                    ++cnt;
                    link(fromRow[i][j], nCnt, used, cst[i][j][0] * -1);
                    link(fromRow[i][j], nCnt, numeric_limits<int>::max(), cst[i][j][0]);
                }
                int used = abs(a[orii][j][0] - cnt);
                curCost += used * cst[orii][j][0];
                if (a[orii][j][0] > cnt)
                    link(nCnt, T, used, cst[orii][j][0] * -1);
                link(nCnt, T, numeric_limits<int>::max(), cst[orii][j][0]);
                --i;
            }
    LOG("%lld %lld\n", curCost.d, curCost.infs);
    curCost += mcf();
    LOG("%lld %lld\n", curCost.d, curCost.infs);
    if (curCost.infs)
        puts("-1");
    else
        printf("%lld\n", curCost.d);
    return 0;
}
Num mcf() {
    Num cost = 0;
    while (spth()) {
        LOG("%lld %lld\n", dist[T].d, dist[T].infs);
        if (!(dist[T] < 0))
            break;
        int flow = numeric_limits<int>::max();
        int u = T;
        while (u != S) {
            flow = min(flow, pth[u]->rev->w);
            u = pth[u]->v;
        }
        cost += dist[T] * flow;
        u = T;
        while (u != S) {
            pth[u]->w += flow;
            pth[u]->rev->w -= flow;
            u = pth[u]->v;
        }
    }
    return cost;
}
bool spth() {
    queue<int> q;
    memset(pth, 0, sizeof(pth));
    memset(inq, 0, sizeof(inq));
    inq[S] = true;
    dist[S] = 0;
    q.push(S);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        LOG("BFS %d %lld %lld\n", u, dist[u].d, dist[u].infs);
        inq[u] = false;
        for (Edge* p = g[u]; p; p = p->next)
            if (p->w && p->v != S && (!pth[p->v] || dist[u] + p->c < dist[p->v])) {
                dist[p->v] = dist[u] + p->c;
                pth[p->v] = p->rev;
                if (!inq[p->v]) {
                    q.push(p->v);
                    inq[p->v] = true;
                }
            }
    }
    return pth[T];
}
void link(int u, int v, int w, const Num& c) {
    LOG("LINK %d %d %d %lld + %lldINF\n", u, v, w, c.d, c.infs);
    Edge *p = addEdge(u, v, w, c), *q = addEdge(v, u, 0, c * -1);
    p->rev = q;
    q->rev = p;
}
Edge* addEdge(int u, int v, int w, const Num& c) {
    static Edge pool[V * V * 4];
    static Edge* p = pool;
    p->v = v;
    p->w = w;
    p->c = c;
    p->next = g[u];
    g[u] = p;
    return p++;
}

T2 Cross sum

可以看成二分图上环的约束，找一颗 DFS 树，随机化。

#include 
#include 
#include 
#include 
#include 
#define LOG(FMT...)  // fprintf(stderr, FMT)
using namespace std;
typedef long long ll;
struct Edge {
    int v;
    ll w;
    bool vis, tree;
    Edge *next, *rev;
};
const int N = 210, M = N * N;
int t, n, m;
bool gg;
int typ[N][N], rowu[N][N], colu[N][N];
int coli[M], rowi[M];
bool vis[M];
ll wt[M];
ll ans[N][N];
Edge* g[M];
Edge pool[M * 2];
Edge* ptop = pool;
set<ll> hsh;
ll rnd();
Edge* addEdge(int u, int v);
void link(int u, int v);
void dfs(int u);
void dfs2(int u);
int main() {
    srand(time(NULL));
    //    freopen("test.in", "r", stdin);
    scanf("%d", &t);
    while (t--) {
        ptop = pool;
        gg = false;
        hsh.clear();
        memset(g, 0, sizeof(g));
        memset(vis, 0, sizeof(vis));
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) scanf("%d", &typ[i][j]);
        int vc = 0;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) {
                if (typ[i][j] & 1) {
                    scanf("%lld", &wt[++vc]);
                    coli[vc] = j;
                    rowi[vc] = 0;
                    for (int k = i + 1; typ[k][j] == 4; ++k) colu[k][j] = vc;
                }
                if (typ[i][j] & 2) {
                    scanf("%lld", &wt[++vc]);
                    rowi[vc] = i;
                    for (int k = j + 1; typ[i][k] == 4; ++k) rowu[i][k] = vc;
                }
            }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j)
                if (typ[i][j] == 4)
                    link(rowu[i][j], colu[i][j]);
        for (int i = 1; i <= vc; ++i)
            if (!vis[i]) {
                dfs(i);
                dfs2(i);
                if (wt[i]) {
                    gg = true;
                    LOG("FAIL %d %lld\n", i, wt[i]);
                }
            }
        if (gg) {
            printf("-1\n");
            continue;
        }
        for (int i = 1; i <= vc; ++i)
            if (rowi[i])
                for (Edge* p = g[i]; p; p = p->next) ans[rowi[i]][coli[p->v]] = p->w;
        for (int i = 1; i <= n; ++i) {
            bool flag = false;
            for (int j = 1; j <= m; ++j)
                if (typ[i][j] == 4) {
                    if (!flag)
                        printf("%lld", ans[i][j]);
                    else
                        printf(" %lld", ans[i][j]);
                    flag = true;
                }
            if (flag)
                printf("\n");
        }
    }
    return 0;
}
void dfs2(int u) {
    for (Edge* p = g[u]; p; p = p->next)
        if (p->tree) {
            p->tree = p->rev->tree = false;
            dfs2(p->v);
            p->w = p->rev->w = wt[p->v];
            if (hsh.count(p->w) || p->w == 0) {
                LOG("CRASH AT (%d, %d), %lld\n", u, p->v, p->w);
                gg = true;
            }
            hsh.insert(p->w);
            wt[u] ^= p->w;
        }
}
void dfs(int u) {
    vis[u] = true;
    for (Edge* p = g[u]; p; p = p->next)
        if (!p->vis)
            if (vis[p->v]) {
                p->vis = p->rev->vis = true;
                p->w = p->rev->w = rnd();
                LOG("CHOOSE %d %d %lld\n", u, p->v, p->w);
                wt[u] ^= p->w;
                wt[p->v] ^= p->w;
                hsh.insert(p->w);
            } else {
                p->vis = p->rev->vis = true;
                p->tree = p->rev->tree = true;
                dfs(p->v);
            }
}
Edge* addEdge(int u, int v) {
    Edge* p = ptop;
    ++ptop;
    p->v = v;
    p->vis = false;
    p->tree = false;
    p->next = g[u];
    g[u] = p;
    return p;
}
void link(int u, int v) {
    Edge *p = addEdge(u, v), *q = addEdge(v, u);
    p->rev = q;
    q->rev = p;
    LOG("%d(%lld) LINK %d(%lld)\n", u, wt[u], v, wt[v]);
}
ll rnd() { return rand() | ((ll)rand() << 16) | ((ll)rand() << 28); }

T3 Kreuzsummen

经过分类讨论的体力活，候选集合必然在 $0\sim 2$ 个区间以内，于是问题规约为三维数点。$\Theta (n{\log }^{2}n)$

#include 
#include 
#include 
#include 
#include 
#include 
#include 

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#ifdef LBT

#define LOG(FMT...) fprintf(stderr, FMT)

#else

#define LOG(FMT...)

#endif

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

typedef pair<int, int> Range;
typedef vector<Range> Ranges;

// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define L first
#define R second

const int N = 100010;

int n, m, k, cnt;

int fw[N];
int res[N * 4];

int lowBit(int x) { return x & -x; }

struct C2D {
  vector<tuple<int, int, int>> modi[N];
  int r, c;

  void ins(int x, int y1, int y2, int v) {
    for (; x <= r; x += lowBit(x)) {
      modi[x].emplace_back(0, y1, v);
      modi[x].emplace_back(0, y2 + 1, -v);
    }
  }

  void qry(int x1, int x2, int y) {
    ++cnt;
    for (; x2; x2 -= lowBit(x2))
      modi[x2].emplace_back(1, y, cnt);
    --x1;
    for (; x1; x1 -= lowBit(x1))
      modi[x1].emplace_back(-1, y, cnt);
  }

  void run() {
    for (int i = 1; i <= r; ++i) {
      for (const auto& tup : modi[i]) {
        int typ, x, k;
        tie(typ, k, x) = tup;
        if (typ == 0) {
          for (; k <= c; k += lowBit(k))
            fw[k] += x;
        } else
          for (; k; k -= lowBit(k))
            res[x] += typ * fw[k];
      }
      for (const auto& tup : modi[i]) {
        int typ, x, k;
        tie(typ, k, x) = tup;
        if (typ == 0)
          for (; k <= c; k += lowBit(k))
            fw[k] = 0;
      }
    }
  }
} r, c;

vector<tuple<int, int, int, int>> md[N];

Ranges cand(int l, ll s) {
  if (l == 1) {
    if (s <= k) return {Range(s, s)};
    return {};
  }
  if (k < l) return {};
  s -= l * (ll)(l + 1) / 2;
  if (s < 0 || s > (k - l) * (ll)l)
    return {};
  if (k == l) return {Range(1, k)};
  if (s == (k - l) * (ll)l) return {Range(k - l + 1, k)};
  if (s == 0) return {Range(1, l)};
  if (l == 2) {
    s += 3;
    int my = max((int)s - k, 1);
    if (s % 2 == 0) {
      if (my <= s / 2 - 1)
        return {Range(my, s / 2 - 1), Range(s / 2 + 1, s - my)};
      return {};
    }
    return {Range(my, s - my)};
  }
  if (k - l == 1)
    return {Range(1, k - s - 1), Range(k - s + 1, k)};
  int pos = s / (k - l);
  if (pos == 0) {
    if (s % (k - l) == 1)
      return {Range(1, l - 1), Range(l + 1, l + 1)};
    return {Range(1, l + s % (k - l))};
  }
  if (pos == l - 1) {
    if (s % (k - l) == k - l - 1)
      return {Range(k - l, k - l), Range(k - l + 2, k)};
    return {Range(1 + (s % (k - l)), k)};
  }
  return {Range(1, k)};
}

int main() {
#ifdef LBT
  freopen("test.in", "r", stdin);
  int nol_cl = clock();
#endif

  int tt;
  scanf("%d%d%d%d", &n, &m, &k, &tt);
  while (tt--) {
    int t, x, y1, y2;
    ll s;
    scanf("%d%d%d%d%lld", &t, &x, &y1, &y2, &s);
    for (const Range& rg : cand(y2 - y1 + 1, s)) {
      md[rg.L - 1].emplace_back(t, -x, y1, y2);
      md[rg.R].emplace_back(t, x, y1, y2);
    }
  }
  r.r = n; r.c = m;
  c.r = m; c.c = n;
  for (int i = k; i >= 0; --i)
    for (const auto& tup : md[i]) {
      int t, x, y1, y2;
      tie(t, x, y1, y2) = tup;
      if (t == 0) {
        if (x > 0) {
          r.ins(x, y1, y2, 1);
          c.qry(y1, y2, x);
        } else {
          r.ins(-x, y1, y2, -1);
          c.qry(y1, y2, -x);
        }
      } else {
        if (x > 0) {
          c.ins(x, y1, y2, 1);
          r.qry(y1, y2, x);
        } else {
          c.ins(-x, y1, y2, -1);
          r.qry(y1, y2, -x);
        }
      }
    }
  r.run();
  c.run();
  ll ans = 0, cur = 0;
  cnt = 0;
  for (int i = k; i >= 0; --i) {
    for (const auto &tup : md[i]) {
      int t, x, y1, y2;
      tie(t, x, y1, y2) = tup;
      if (x > 0)
        cur += res[++cnt];
      else
        cur -= res[++cnt];
    }
    ans += cur;
  }
  printf("%lld\n", ans);

#ifdef LBT
  LOG("Time: %dms\n", int ((clock()
          -nol_cl) / (double)CLOCKS_PER_SEC * 1000));
#endif
  return 0;
}