Description

給出一個 \(n*n\) 的矩陣,每一格有一個非負整數 \(A_{i,j}\) ,(\(A_{i,j} <= 1000\))現在從 \((1,1)\) 出發,可以往右或者往下走,最後到達 \((n,n)\) ,每達到一格,把該格子的數取出來,該格子的數就變成 \(0\) ,這樣一共走 \(K\) 次,現在要求 \(K\) 次所達到的方格的數的和最大

Solution

一條邊 \((a,b)\) 表示容量為 \(a\) ，費用為 \(b\) 。
把每個點拆成兩個點，入點和出點。入點用來接受邊，出點用來發出邊
源點向 \((1,1)\) 連一條邊 \((k,0)\) ，\((n,n)\)

Code

#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int N = 550; 
int n, m, cnt, vis[N * N * 3], dis[N * N * 3]; 
int S, T, k, pre[N * N * 3], f[N * N * 3];  
struct node {
  int d, sid, tid; 
}a[N][N];
struct edge {
  int v, w, f; edge *next, *rev;
}pool[N * N * 2], *head[N * N * 3], *r[N * N * 3]; 
inline void addedge(int u, int v, int f, int w) {
  edge *p = &pool[++cnt], *q = &pool[++cnt]; 
  p->v = v, p->f = f, p->w = w,  p->next = head[u], head[u] = p; p->rev = q; 
  q->v = u, q->f = 0, q->w = -w, q->next = head[v], head[v] = q; q->rev = p;
}
inline bool spfa() {
  for(int i = S; i <= T; i++) r[i] = NULL, dis[i] = INF, vis[i] = 0, pre[i] = -1; 
  queue <int> Q; Q.push(S); vis[S] = 1; dis[S] = 0; f[S] = INF; 
  while(!Q.empty()) {
    int u = Q.front(); Q.pop(); vis[u] = 0; 
    for(edge *p = head[u]; p; p = p->next) {
      int v = p->v; 
      if(p->f > 0 && dis[v] > dis[u] + p->w) {
        dis[v] = dis[u] + p->w;
        pre[v] = u, r[v] = p; 
        f[v] = min(f[u], p->f); 
        if(!vis[v]) vis[v] = 1, Q.push(v); 
      }
    }
  } 
  return pre[T] != -1; 
}
inline int MCMF() {
  int ans = 0;  
  while(spfa()) {
    for(int i = T; i != S; i = pre[i]) {
      r[i]->f -= f[T]; r[i]->rev->f += f[T];  
    } ans += dis[T] * f[T];
  } return ans; 
}
int main() { 
  scanf("%d %d", &n, &k); S = 0, T = 2 * n * n + 1; 
  addedge(S, 1, k, 0); 
  for(int i = 1; i <= n; i++) {
    for(int j = 1; j <= n; j++) {
      int x; scanf("%d", &x); 
      int id = (i - 1) * n + j; 
      a[i][j].sid = 2 * id - 1;
      a[i][j].tid = 2 * id; 
      addedge(a[i][j].sid, a[i][j].tid, 1, -x); 
      addedge(a[i][j].sid, a[i][j].tid, INF, 0); 
    }
  } 
  for(int i = 1; i <= n; i++)
    for(int j = 1; j <= n; j++) {
      if(i < n) addedge(a[i][j].tid, a[i + 1][j].sid, INF, 0);
      if(j < n) addedge(a[i][j].tid, a[i][j + 1].sid, INF, 0); 
    }
  addedge(a[n][n].tid, T, k, 0); 
  printf("%d\n", -MCMF());
  return 0; 
}