POJ3422 Kaka's Matrix Travels

红太狼 2022-07-21 11:13 16阅读 0赞

Kaka's Matrix Travels

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;9244</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;3736</td> 
  </tr> 
 </tbody> 
</table>

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 *K*th 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 *K*th travel.

Sample Input

3 2
    1 2 3
    0 2 1
    1 4 2

Sample Output

Source

[POJ Monthly--2007.10.06][], Huang, Jinsong

测试一下最小费用最大流的模板。很经典的建图方法。

#include <algorithm>
    #include <iostream>
    using namespace std;
    
    const int MAXN = 10000;
    const int MAXM = 100000;
    const int INF = 0x3f3f3f3f;
    struct Edge
    {
        int to,next,cap,flow,cost;
    }edge[MAXM];
    int head[MAXN],tol;
    int pre[MAXN],dis[MAXN];
    bool vis[MAXN];
    int N;//节点总个数，节点编号从0~N-1
    void init(int n)
    {
        N = n;
        tol = 0;
        memset(head,-1,sizeof(head));
    }
    void addedge(int u,int v,int cap,int cost)
    {
        edge[tol].to = v;
        edge[tol].cap = cap;
        edge[tol].cost = cost;
        edge[tol].flow = 0;
        edge[tol].next = head[u];
        head[u] = tol++;
        edge[tol].to = u;
        edge[tol].cap = 0;
        edge[tol].cost = -cost;
        edge[tol].flow = 0;
        edge[tol].next = head[v];
        head[v] = tol++;
    }
    bool spfa(int s,int t)
    {
        queue<int>q;
        for(int i = 0;i < N;i++)
        {
            dis[i] = INF;
            vis[i] = false;
            pre[i] = -1;
        }
        dis[s] = 0;
        vis[s] = true;
        q.push(s);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            vis[u] = false;
            for(int i = head[u]; i != -1;i = edge[i].next)
            {
                int v = edge[i].to;
                if(edge[i].cap > edge[i].flow &&
                   dis[v] > dis[u] + edge[i].cost )
                {
                    dis[v] = dis[u] + edge[i].cost;
                    pre[v] = i;
                    if(!vis[v])
                    {
                        vis[v] = true;
                        q.push(v);
                    }
                }
            }
        }
        if(pre[t] == -1)return false;
        else return true;
    }
    //返回的是最大流，cost存的是最小费用
    int minCostMaxflow(int s,int t,int &cost)
    {
        int flow = 0;
        cost = 0;
        while(spfa(s,t))
        {
            int Min = INF;
            for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
            {
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            }
            for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
                cost += edge[i].cost * Min;
            }
            flow += Min;
        }
        return flow;
    }
    
    int a[55][55];
    int main()
    {
        int n,k;
        while(scanf("%d%d",&n,&k) == 2)
        {
            for(int i = 0;i < n;i++)
                for(int j = 0;j < n;j++)
                    scanf("%d",&a[i][j]);
            init(2*n*n+2);
            for(int i = 0;i < n;i++)
                for(int j = 0;j < n;j++)
                {
                    addedge(n*i+j+1,n*n+n*i+j+1,1,-a[i][j]);
                    addedge(n*i+j+1,n*n+n*i+j+1,INF,0);
                }
    
            for(int i = 0;i < n;i++)
                for(int j = 0;j < n;j++)
                {
                    if(i < n-1)
                        addedge(n*n+n*i+j+1,n*(i+1)+j+1,INF,0);
                    if(j < n-1)
                        addedge(n*n+n*i+j+1,n*i+j+1+1,INF,0);
                }
            addedge(0,1,k,0);
            addedge(2*n*n,2*n*n+1,INF,0);
            int cost;
            minCostMaxflow(0,2*n*n+1,cost);
            printf("%d\n",-cost);
        }
        return 0;
    }

[POJ Monthly--2007.10.06]: http://poj.org/searchproblem?field=source&key=POJ+Monthly--2007.10.06