题目链接：点击打开链接
思路：
我们用流量来限制走的次数， 因为权值在结点上， 我们考虑拆点， 因为每个点的权值只能获得一次， 我们对于每个点连两条边， 一条容量为1费用为权值， 一条容量为k-1，费用为0， 跑最大费用流就行了。
细节参见代码：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include <cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; typedef long long ll; typedef long double ld; const double eps = 1e-6; const double PI = acos(-1); const int mod = 1000000000 + 7; const int INF = 0x3f3f3f3f; const int seed = 131; const ll INF64 = ll(1e18); const int maxn = 50*50*2 + 10; int T,n,m, k; struct Edge { int from, to, cap, flow, cost; }; struct MCMF { int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; int inq[maxn]; // 是否在队列中 int d[maxn]; // Bellman-Ford int p[maxn]; // 上一条弧 int a[maxn]; // 可改进量 void init(int n) { this->n = n; for(int i = 0; i < n; i++) G[i].clear(); edges.clear(); } 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}); m = edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BellmanFord(int s, int t, int& flow, int& cost) { for(int i = 0; i < n; i++) d[i] = INF; memset(inq, 0, sizeof(inq)); d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF; 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[e.to] > d[u] + e.cost) { 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 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]].from; } return true; } int Mincost(int s, int t) { int cost = 0, flow = 0; while(BellmanFord(s, t, flow, cost)); return cost; } }; MCMF g; int a[55][55]; inline int get_id(int r, int c) { return (r-1)*n + c; } int main() { while(~scanf("%d%d",&n, &k)) { for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); } } int src = 0, stc = 2*n*n + 5; g.init(2*n*n + 8); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { int id = get_id(i, j); if(k) { g.AddEdge(id, id+n*n, 1, -a[i][j]); g.AddEdge(id, id+n*n, k-1, 0); } if(i == 1 && j == 1) { g.AddEdge(src, id, INF, 0); } else if(i == n && j == n) { g.AddEdge(id+n*n, stc, INF, 0); continue; } if(i < n) g.AddEdge(id + n*n, get_id(i+1, j), k, 0); if(j < n) g.AddEdge(id + n*n, get_id(i, j+1), k, 0); } } printf("%d\n", -g.Mincost(src, stc)); } return 0; }
