Poj 3422（最小费用流）

Kaka's Matrix Travels

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6970 Accepted: 2759

Description

On an N × N chessboard with a non-negative number in each grid, Kaka starts his matrix travels with SUM = 0. For each travel, Kaka moves one rook from the left-upper grid to the right-bottom one, taking care that the rook moves only to the right or down. Kaka adds the number to SUM in each grid the rook visited, and replaces it with zero. It is not difficult to know the maximum SUM Kaka can obtain for his first travel. Now Kaka is wondering what is the maximum SUM he can obtain after his Kth travel. Note the SUM is accumulative during the K travels.

Input

The first line contains two integers N and K (1 ≤ N ≤ 50, 0 ≤ K ≤ 10) described above. The following N lines represents the matrix. You can assume the numbers in the matrix are no more than 1000.

Output

The maximum SUM Kaka can obtain after his Kth travel.

Sample Input

3 21 2 30 2 11 4 2 

Sample Output

15

Source

POJ Monthly--2007.10.06, Huang, Jinsong

给一个矩阵，每个点有个权值，从左上角走到到右下角，求路径上所有权值的最大和。如果只能走一条路，那就是个经典的dp问题，这道题关键是可以走多条路径，并且某个点走过后，它的权值就变为0了。如果不是放在网络流中，还真是想不到用网络流去做，不过知道后就好办了。把每个点拆成两个点，分别叫”出点“，和“入点”，然后入点连向出点，权值为该点的负权值，容量为1。因为每个点只能往右或下走，这样从出点分别连向相邻点的出点和入点，容量为无穷，费用是0。最后跑一遍限定流量的最大流就可以了。

#include <cstdio>#include <cstring>#include <vector>#include<iostream>#include <algorithm>#include<queue>using namespace std;const int maxn = 5000 + 5; // 顶点的最大数目const int INF = 1000000000;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,单位流量的费用  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;//s-t不连通，失败退出    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;  }  // 需要保证初始网络中没有负权圈  int Mincost(int s, int t) {    int flow = 0,cost = 0;    while(BellmanFord(s, t,flow, cost));    return cost;  }};MCMF g;int M[maxn][maxn];int main(){    int n,k;    while(scanf("%d%d",&n,&k) != EOF){        for(int i = 0;i < n;i++){            for(int j = 0;j < n;j++){                scanf("%d",&M[i][j]);            }        }        g.init(2*n*n+1);        int sorce = 2*n*n,sink = 2*n*n-1;        g.AddEdge(sorce,n*n,k,0);        for(int i = 0;i < n;i++){            for(int j = 0;j < n;j++){                g.AddEdge(i*n+j,i*n+j+n*n,1,-M[i][j]);            }        }        for(int i = 0;i < n;i++){            for(int j = 0;j < n;j++){                if(j != n-1){                    g.AddEdge(i*n+j+n*n,i*n+j+1,INF,0);                    g.AddEdge(i*n+j+n*n,i*n+j+1+n*n,INF,0);                }                if(i != n-1){                    g.AddEdge(i*n+j+n*n,(i+1)*n+j,INF,0);                    g.AddEdge(i*n+j+n*n,(i+1)*n+j+n*n,INF,0);                }            }        }        int ans = -g.Mincost(sorce,sink);        if(k != 0) ans += M[0][0];        printf("%d\n",ans);    }    return 0;}