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AtCoder Beginner Contest 225 G X

发布时间 2023-06-05 22:23:31作者: zltzlt

洛谷传送门

AtCoder 传送门

感觉是一种很新的建图方式。

首先肯定是最小割没跑了。尝试描述损失。

也不难发现要连 \(S \to (i, j)\)，容量 \(a_{i, j}\)，这样描述出了每个点选与不选的状态。

考虑如何描述 X。考虑连边 \((i, j) \to (i + 1, j - 1), (i, j) \to (i + 1, j + 1), (n, 1) \to T, (n, m) \to T\)，容量均为 \(C\)。表示对于一条斜线，在它的终止位置的连向下一个点的边会被割断，否则 \(S\) 能到达这些点跟下一个斜线的点，那跟选终止位置下一个点的效果相同，还不如不割，即选择它。

复杂度好像是错的，但是可过。

code 
// Problem: G - X
// Contest: AtCoder - UNICORN Programming Contest 2021(AtCoder Beginner Contest 225)
// URL: https://atcoder.jp/contests/abc225/tasks/abc225_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 110;
const int maxm = 1000100;
const ll inf = 0x3f3f3f3f3f3f3f3fLL;

ll n, m, K, a[maxn][maxn], head[maxm], len = 1, ntot, S, T, id[maxn][maxn];
struct edge {
	ll to, next, cap, flow;
} edges[maxm];

inline void add_edge(ll u, ll v, ll c, ll f) {
	edges[++len].to = v;
	edges[len].next = head[u];
	edges[len].cap = c;
	edges[len].flow = f;
	head[u] = len;
}

struct Dinic {
	ll d[maxm], cur[maxm];
	bool vis[maxm];
	
	inline void add(ll u, ll v, ll c) {
		add_edge(u, v, c, 0);
		add_edge(v, u, 0, 0);
	}
	
	inline bool bfs() {
		for (int i = 1; i <= ntot; ++i) {
			d[i] = -1;
			vis[i] = 0;
		}
		queue<int> q;
		q.push(S);
		vis[S] = 1;
		d[S] = 0;
		while (q.size()) {
			int u = q.front();
			q.pop();
			for (int i = head[u]; i; i = edges[i].next) {
				edge &e = edges[i];
				if (!vis[e.to] && e.cap > e.flow) {
					vis[e.to] = 1;
					d[e.to] = d[u] + 1;
					q.push(e.to);
				}
			}
		}
		return vis[T];
	}
	
	ll dfs(ll u, ll a) {
		if (u == T || !a) {
			return a;
		}
		ll flow = 0, f;
		for (ll &i = cur[u]; i; i = edges[i].next) {
			edge &e = edges[i];
			if (d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i ^ 1].flow -= f;
				flow += f;
				a -= f;
				if (!a) {
					break;
				}
			}
		}
		return flow;
	}
	
	inline ll solve() {
		ll flow = 0;
		while (bfs()) {
			for (int i = 1; i <= ntot; ++i) {
				cur[i] = head[i];
			}
			flow += dfs(S, inf);
		}
		return flow;
	}
} solver;

void solve() {
	scanf("%lld%lld%lld", &n, &m, &K);
	S = ++ntot;
	T = ++ntot;
	ll s = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			scanf("%lld", &a[i][j]);
			id[i][j] = ++ntot;
			s += a[i][j];
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			solver.add(S, id[i][j], a[i][j]);
			if (i < n && j > 1) {
				solver.add(id[i][j], id[i + 1][j - 1], K);
			} else {
				solver.add(id[i][j], T, K);
			}
			if (i < n && j < m) {
				solver.add(id[i][j], id[i + 1][j + 1], K);
			} else {
				solver.add(id[i][j], T, K);
			}
		}
	}
	printf("%lld\n", s - solver.solve());
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}