因为每个格子可以走无数次，但走过一次之后数字就变成了0，也就是只有一次可以加上格子里的数字。所以要拆点（X->Xi,Xj），在Xi和Xj之间连一条容量为1，费用为数字的相反数的边（取走该格数字），再连一条容量为INF，费用为0的边（第2,3...n次走该格）。再从每个格子的Xj点向左面及下面的格子的Xi点连一条容量为1，费用为0的边，最后从源点向第一个格子的Xi连一条容量为k（走k次）费用为0，从最后一个点的Xj向汇点连一条同样的边。求解最小费用最大流，费用的相反数就是答案。

总时间耗费: 149ms 
总内存耗费: 1 kB

#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std; const int INF = 1e8 + 7;
const int maxn = 50 * 50 * 2 + 10;struct Edge{int from, to, cap, flow, cost;
};vector<Edge> edges;
vector<int> G[maxn];
int n, k, s, t;void encode(int i, int j, int& THIS, int& NEXT, int& DOWN, int& LEFT) {i--; j--;THIS = i * n + j + 1;NEXT = THIS + n*n;DOWN = i < n-1 ? THIS + n : 0;LEFT = j < n-1 ? THIS + 1 : 0;
}void AddEdge(int from, int to, int cap, int cost) {edges.push_back((Edge){from, to, cap, 0, cost});edges.push_back((Edge){to, from, 0, 0, -cost});int m = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);
}int d[maxn], a[maxn], p[maxn];
bool inq[maxn];bool SPFA(int &cost) {for(int i = 0; i <= t; i++) d[i] = INF;memset(inq, 0, sizeof(inq));d[s] = 0; inq[s] = 1; a[s] = INF; p[s] = 0;queue<int> Q;Q.push(s);while(!Q.empty()) {int u = Q.front(); Q.pop();inq[u] = 0;for(int i = 0; i < G[u].size(); i++) {Edge& e = edges[G[u][i]];if(e.cap > e.flow && d[u] + e.cost < d[e.to]) {d[e.to] = d[u] + e.cost;p[e.to] = G[u][i];a[e.to] = min(a[u], e.cap - e.flow);if(!inq[e.to]) {Q.push(e.to);inq[e.to] = 1;}}}}if(d[t] == INF) return 0;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]].from;}return 1;
}int maxflow() {int cost = 0;while(SPFA(cost));return cost;
}int main() {cin >> n >> k;s = 0; t = n*n*2 + 1;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++) {int THIS, NEXT, DOWN, LEFT;encode(i, j, THIS, NEXT, DOWN, LEFT);int w;cin >> w;AddEdge(THIS, NEXT, 1, -w);AddEdge(THIS, NEXT, INF, 0);if(LEFT) AddEdge(NEXT, LEFT, INF, 0);if(DOWN) AddEdge(NEXT, DOWN, INF, 0);}AddEdge(s, 1, k, 0);AddEdge(t-1, t, k, 0);cout << -maxflow() << endl;return 0;
}