【BZOJ5251】【2018多省省队联测】劈配

【题目链接】

• 点击打开链接
  

【思路要点】

• 按照题意建图，跑网络流即可。
• 由于\(C\)的限制，图中点数为\(O(N+M)\)级别，边数为\(O(N*C)\)级别的。
• 并且，每一次运行Dinic算法只会增广一条路径，因此，时间复杂度为\(O(T*C*N^3)\)(\(N\)、\(M\)同阶)。

【代码】

#include
using namespace std;
const int MAXP = 505;
const int INF = 1e9;
template  void read(T &x) {
	int f = 1; x = 0;
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
	for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
	x *= f;
}
struct edge {int dest, flow; unsigned home; };
int T, C;
int n, m, s, t, tot, flow;
int goal[MAXP];
int dist[MAXP];
int ans[MAXP], bns[MAXP];
vector  a[MAXP];
vector  b[MAXP][MAXP];
unsigned sz[MAXP], curr[MAXP];
int dinic(int pos, int limit) {
	if (pos == t) return limit;
	int used = 0, tmp;
	for (unsigned &i = curr[pos]; i < a[pos].size(); i++)
		if (a[pos][i].flow && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)))) {
			used += tmp;
			a[pos][i].flow -= tmp;
			a[a[pos][i].dest][a[pos][i].home].flow += tmp;
			if (used == limit) return used;
		}
	return used;
}
bool bfs() {
	static int q[MAXP], l = 0, r = -1;
	for (int i = 0; i <= r; i++)
		dist[q[i]] = 0;
	q[l = r = 0] = s, dist[s] = 1;
	while (l <= r) {
		int tmp = q[l++];
		for (unsigned i = 0; i < a[tmp].size(); i++)
			if (a[tmp][i].flow && dist[a[tmp][i].dest] == 0) {
				q[++r] = a[tmp][i].dest;
				dist[a[tmp][i].dest] = dist[tmp] + 1;
			}
	}
	return dist[t] != 0;
}
void packup() {
	for (int i = 1; i <= tot; i++)
		sz[i] = a[i].size();
}
void restore() {
	for (int i = 1; i <= tot; i++)
		a[i].resize(sz[i]);
}
void addedge(int s, int t, int flow) {
	a[s].push_back((edge) {t, flow, a[t].size()});
	a[t].push_back((edge) {s, 0, a[s].size() - 1});
}
int main() {
	read(T), read(C);
	while (T--) {
		while (tot) {
			a[tot].clear();
			tot--;
		}
		s = ++tot;
		read(n), read(m);
		for (int i = 1; i <= n; i++)
			addedge(s, ++tot, 1);
		t = n + m + 2;
		for (int i = 1; i <= m; i++) {
			int x; read(x);
			addedge(++tot, t, x);
		}
		++tot;
		for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			int x; read(x);
			if (x != 0) b[i][x].push_back(j);
		}
		for (int i = 1; i <= n; i++)
			read(goal[i]);
		for (int i = 1; i <= n; i++)
			ans[i] = m + 1;
		memset(bns, -1, sizeof(bns));
		packup();
		for (int j = 1; j <= n; j++) {
			if (bns[j] != -1) continue;
			for (int val = 1; val <= goal[j]; val++)
			for (unsigned k = 0; k < b[j][val].size(); k++)
				addedge(j + 1, n + 1 + b[j][val][k], 1);
			if (!bfs()) bns[j] = j;
			restore();
		}
		for (int i = 1; i <= n; i++) {
			packup();
			for (int j = 1; j <= m; j++) {
				for (unsigned k = 0; k < b[i][j].size(); k++)
					addedge(i + 1, n + 1 + b[i][j][k], 1);
				if (bfs()) {
					ans[i] = j;
					memset(curr, 0, sizeof(curr));
					flow += dinic(s, INF);
					break;
				} else restore();
			}
			if (bns[i] == -1) {
				if (ans[i] <= goal[i]) bns[i] = 0;
				else bns[i] = 1;
			}
			packup();
			for (int j = i + 1; j <= n; j++) {
				if (bns[j] != -1) continue;
				for (int val = 1; val <= goal[j]; val++)
				for (unsigned k = 0; k < b[j][val].size(); k++)
					addedge(j + 1, n + 1 + b[j][val][k], 1);
				if (!bfs()) bns[j] = j - i;
				restore();
			}
		}
		for (int i = 1; i <= n; i++)
			printf("%d ", ans[i]);
		printf("\n");
		for (int i = 1; i <= n; i++)
			printf("%d ", bns[i]);
		printf("\n");
		for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			b[i][j].clear();
	}
	return 0;
}