HDU 3376 Matrix Again(费用流)
2015-09-06 00:05
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题目大意:给出一张网格,网格上面有相应的正数,现在要求你从左上角出发,走到右下角,再从右下角出发,走回左上角,沿路经过的格子不能重复(除了左上角和右下角的格子),且从左上角走到右下角时只能往右或者往下走,从右下角走到左上角只能往上或者往左。现在问所走的格子的最大总和是多少
解题思路:遵守规则的话,从左上角到右下角,和从右下角到左上角,其实走的路径是对称的,能从上往下走,就表示能从下往上走,所以我们以左上角为源点,右下角为汇点建图
因为每个格子(除了左上角和右下角)只能走一次,所以拆点,容量为1,费用取格子的数的相反数,这样就能保证只走一次了
两个特殊的格子特殊考虑,容量为2,费用还是取相反数
这样,只要流了2的流量,就表示任务完成了,最后的最小费用取反,再减去重复走的两个特殊格子的费用
解题思路:遵守规则的话,从左上角到右下角,和从右下角到左上角,其实走的路径是对称的,能从上往下走,就表示能从下往上走,所以我们以左上角为源点,右下角为汇点建图
因为每个格子(除了左上角和右下角)只能走一次,所以拆点,容量为1,费用取格子的数的相反数,这样就能保证只走一次了
两个特殊的格子特殊考虑,容量为2,费用还是取相反数
这样,只要流了2的流量,就表示任务完成了,最后的最小费用取反,再减去重复走的两个特殊格子的费用
[code]#include <cstdio> #include <cstring> #include <queue> #include <vector> #include <algorithm> using namespace std; const int MAXNODE = 600 * 600 * 2 + 10; const int MAXEDGE = 4 * MAXNODE; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge{ int u, v, next; Type cap, flow, cost; Edge() {} Edge(int u, int v, Type cap, Type flow, Type cost, int next): u(u), v(v), cap(cap), flow(flow), cost(cost), next(next) {} }; struct MCMF{ int n, m, s, t; Edge edges[MAXEDGE]; int head[MAXNODE]; int p[MAXNODE]; Type d[MAXNODE]; Type a[MAXNODE]; bool inq[MAXNODE]; int flow, cost; void init(int n) { this->n = n; memset(head, -1, sizeof(head)); m = 0; } void AddEdge(int u, int v, Type cap, Type cost) { edges[m] = Edge(u, v, cap, 0, cost, head[u]); head[u] = m++; edges[m] = Edge(v, u, 0, 0, -cost, head[v]); head[v] = m++; } bool BellmanFord(int s, int t, Type &flow, Type &cost) { for (int i = 0; i < n; i++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = true; p[s] = 0; a[s] = INF; queue<int> Q; Q.push(s); while (!Q.empty()) { int u = Q.front(); Q.pop(); inq[u] = false; for (int i = head[u]; ~i; i = edges[i].next) { 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 += a[t] * d[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) { this->s = s; this->t = t; flow = 0, cost = 0; while (BellmanFord(s, t, flow, cost)); return cost; } }mcmf; #define maxn 610 int map[maxn][maxn]; int dir[2][2] = {{0, 1}, {1, 0}}; int n; void init() { for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) scanf("%d", &map[i][j]); } void solve() { int source = 0, sink = n * n * 2 + 1; int t = n * n; mcmf.init(sink + 1); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { if (i == j && (i == 1 || i == n)) mcmf.AddEdge((i - 1) * n + j, (i - 1) * n + j + t, 2, -map[i][j]); else mcmf.AddEdge((i - 1) * n + j, (i - 1) * n + j + t, 1, -map[i][j]); for (int k = 0; k < 2; k++) { int tx = i + dir[k][0]; int ty = j + dir[k][1]; if (tx < 1 || tx > n || ty < 1 || ty > n) continue; mcmf.AddEdge((i - 1) * n + j + t, (tx - 1) * n + ty, 1, 0); } } mcmf.AddEdge(source, 1, 2, 0); mcmf.AddEdge(t * 2, sink, 2, 0); printf("%d\n", -mcmf.MinCost(source, sink) - map[1][1] - map ); } int main() { while (scanf("%d", &n) != EOF) { init(); solve(); } return 0; }
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