HDU 2686 Matrix 3376 Matrix Again（费用流）

HDU 2686 Matrix

题目链接

3376 Matrix Again

题目链接

题意：这两题是一样的，仅仅是数据范围不一样，都是一个矩阵，从左上角走到右下角在从右下角走到左上角能得到最大价值

思路：拆点。建图，然后跑费用流就可以，只是HDU3376这题，极限情况是300W条边，然后卡时间过了2333

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 600 * 600 * 2 + 5;
const int MAXEDGE = 4 * MAXNODE;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow, cost;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow, Type cost) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
		this->cost = cost;
	}
};

struct MCFC {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	int inq[MAXNODE];
	Type d[MAXNODE];
	int p[MAXNODE];
	Type a[MAXNODE];

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}

	void add_Edge(int u, int v, Type cap, Type cost) {
		edges[m] = Edge(u, v, cap, 0, cost);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0, -cost);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bellmanford(int s, int t, Type &flow, Type &cost) {

		for (int i = 0; i < n; i++) d[i] = INF;
		memset(inq, false, sizeof(inq));
		d[s] = 0; inq[s] = true; p[s] = s; a[s] = INF;
		queue<int> Q;
		Q.push(s);
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			inq[u] = false;
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (e.cap > e.flow && d[e.v] > d[u] + e.cost) {
					d[e.v] = d[u] + e.cost;
					p[e.v] = i;
					a[e.v] = min(a[u], e.cap - e.flow);
					if (!inq[e.v]) {Q.push(e.v); inq[e.v] = true;}
				}
			}
		}
		if (d[t] == INF) return false;
		flow += a[t];
		cost += d[t] * a[t];
		int u = t;
		while (u != s) {
			edges[p[u]].flow += a[t];
			edges[p[u]^1].flow -= a[t];
			u = edges[p[u]].u;
		}
		return true;
	}

	Type Mincost(int s, int t) {
		Type flow = 0, cost = 0;
		while (bellmanford(s, t, flow, cost));
		return cost;
	}
} gao;

const int N = 600 * 600 + 5;
const int d[2][2] = {1, 0, 0, 1};

int n, num[N];

int get(int now, int k) {
	int x = now / n;
	int y = now % n;
	x += d[k][0];
	y += d[k][1];
	if (x < 0 || x >= n || y < 0 || y >= n) return -1;
	return x * n + y;
}

int main() {
	while (~scanf("%d", &n)) {
		gao.init(n * n * 2);
		for (int i = 0; i < n * n; i++) {
			scanf("%d", &num[i]);
			if (i == 0) gao.add_Edge(i, i + n * n, 2, -num[i]);
			else if (i == n * n - 1) gao.add_Edge(i, i + n * n, 2, -num[i]);
			else gao.add_Edge(i, i + n * n, 1, -num[i]);
		}
		for (int i = 0; i < n * n; i++) {
			for (int j = 0; j < 2; j++) {
				int next = get(i, j);
				if (next < 0 || next >= n * n) continue;
				gao.add_Edge(i + n * n, next, 2, 0);
			}
		}
		printf("%d\n", -gao.Mincost(0, n * n * 2 - 1) - num[0] - num[n * n - 1]);
	}
	return 0;
}

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