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
#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[n][n]);
}
int main() {
while (scanf("%d", &n) != EOF) {
init();
solve();
}
return 0;
}