HDU 2686 Matrix 3376 Matrix Again（费用流）

最新推荐文章于 2019-09-02 10:57:52 发布

原创最新推荐文章于 2019-09-02 10:57:52 发布 · 948 阅读

0 ·

CC 4.0 BY-SA版权

图论-网络流专栏收录该内容

55 篇文章

订阅专栏

本文探讨了矩阵路径优化问题，通过拆点、建图并应用费用流算法解决从矩阵左上角到右下角及反之的路径选择问题。重点介绍了如何在不同数据范围内优化算法性能，并详细阐述了解决方案的实现过程。

HDU 2686 Matrix

题目链接

3376 Matrix Again

题目链接

题意：这两题是一样的，只是数据范围不一样，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，然后跑费用流即可，不过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;
}