POJ3422 Kaka's Matrix Travels

POJ3422 Kaka's Matrix Travels
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 4905
Accepted: 1910
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 2

1 2 3

0 2 1

1 4 2

Sample Output

15

************************************************************

    ：  n*n   ，         ，           ，          SUM             。    K        SUM。

    ：         。       ，   spfa     。。。         ；              。

  。        ，  x x'        。            ， ，       ：x x'     ，     1，     ，      INF，   0.     ，AC       。

#include <stdio.h>

#include <string.h>

#include <queue>

#define N 50050

#define M 10000005

#define INF 0x3f3f3f3f

#define tk(a,b,c) ((a-1)*n+b+c*n*n)

using namespace std;



int tu[55][55];

int n,k,eid,sink,source;

int head[N],ed[M],fee[M],nxt[M],up[M];

int pre[N],vis[N],dist[N],road[N];



void add_edge(int s,int e,int f,int u)

{

    up[eid]=u;

    ed[eid]=e;         nxt[eid]=head[s];

    fee[eid]=f;        head[s]=eid++;

    up[eid]=0;

    ed[eid]=s;         fee[eid]=-f;

    nxt[eid]=head[e];  head[e]=eid++;

}



int spfa(void)

{

    queue<int>que;

    memset(vis,0,sizeof(vis));

    for(int i=1;i<=sink;i++)

        dist[i]=-1;

    dist[0]=0;

    vis[0]=1;que.push(0);

    while(!que.empty())

    {

        int t=que.front();

        que.pop();

        vis[t]=0;

        for(int i=head[t];~i;i=nxt[i])

        {

            if(up[i]==0)continue;

            int e=ed[i],f=fee[i];

            if(dist[e]<dist[t]+f)

            {

                dist[e]=dist[t]+f;

                road[e]=i;

                pre[e]=t;

                if(!vis[e])

                {

                    vis[e]=1;

                    que.push(e);

                }

            }

        }

    }

    if(pre[sink]!=-1&&dist[sink]>-1)

        return 1;

    return 0;

}



void re(void)

{



    for(int i=1;i<=n;i++)

        for(int j=1;j<=n;j++)

            scanf("%d",&tu[i][j]);

}



void run(void)

{

    memset(head,-1,sizeof(head));

    eid=0;source=0;sink=2*n*n+1;

    for(int i=1;i<=n;i++)

        for(int j=1;j<=n;j++)

        {

            add_edge(tk(i,j,0),tk(i,j,1),tu[i][j],1);

            add_edge(tk(i,j,0),tk(i,j,1),0,k-1);

        }

    for(int i=1;i<=n;i++)

        for(int j=1;j<=n;j++)

        {

            if(n-i)

                add_edge(tk(i,j,1),tk(i+1,j,0),0,k);

            if(n-j)

                add_edge(tk(i,j,1),tk(i,j+1,0),0,k);

        }

    add_edge(source,tk(1,1,0),0,k);

    add_edge(tk(n,n,1),sink,0,k);

    int ans=0;

    int vis[N];

    while(spfa())

        for(int j=sink;j!=source;j=pre[j])

        {

            up[road[j]]-=1;

            up[road[j]^1]+=1;

            ans+=fee[road[j]];

        }

    printf("%d
",ans);

}



int main()

{

    while(scanf("%d%d",&n,&k)==2)

    {

         re();

        run();

    }

    return 0;

}