HDU 2686 Matrix 3376 Matrix Again (cost stream), hdu3376

HDU 2686 Matrix 3376 Matrix Again (cost stream), hdu3376
HDU 2686 Matrix

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3376 Matrix Again

Question Link

These two questions are the same, but the data ranges are different. They are all a matrix. You can get the maximum value from the upper left corner to the lower right corner to the upper left corner.

Idea: Split the points, create a graph, and then run the fee stream. However, the limit of HDU3376 is 2333 sides, and the time is over.

Code:

#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;}