| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9402 | Accepted: 3819 |
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
Source
题意：给出n*n矩阵，每个点上有价值，只能往右or下走一格，只能采一次价值（可以重复走，起点不定）走m次最多的价值。
做法：和
几乎一样==用流量限制m次，所以s->s'流量是m cost=0，能走的j'->i连两条边：cap=1,cost=x;cap=oo,cost=0,1其他边是cap=oo cost=0
/**********
poj3422
768K 204MS C++ 2766B
2016-08-31 11:19:40
***********/
#include <iostream>
#include <cstdio>
using namespace std;
const int oo=1e9;//无穷大
const int maxm=100000;//边的最大数量，为原图的两倍
const int maxn=10000;//点的最大数量
int node,src,dest,edge;//node节点数，src源点，dest汇点，edge边数
int head[maxn],p[maxn],dis[maxn],q[maxn],vis[maxn];//head链表头，p记录可行流上节点对应的反向边，dis计算距离
struct edgenode
{
int to;//边的指向
int flow;//边的容量
int cost;//边的费用
int next;//链表的下一条边
} edges[maxm];
void prepare(int _node,int _src,int _dest);
void addedge(int u,int v,int f,int c);
bool spfa();
inline int min(int a,int b)
{
return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
node=_node;
src=_src;
dest=_dest;
for (int i=0; i<node; i++)
{
head[i]=-1;
vis[i]=false;
}
edge=0;
}
void addedge(int u,int v,int f,int c)
{
edges[edge].flow=f;
edges[edge].cost=c;
edges[edge].to=v;
edges[edge].next=head[u];
head[u]=edge++;
edges[edge].flow=0;
edges[edge].cost=-c;
edges[edge].to=u;
edges[edge].next=head[v];
head[v]=edge++;
}
bool spfa()
{
int i,u,v,l,r=0,tmp;
for (i=0; i<node; i++) dis[i]=oo;
dis[q[r++]=src]=0;
p[src]=p[dest]=-1;
for (l=0; l!=r; ((++l>=maxn)?l=0:1))
{
for (i=head[u=q[l]],vis[u]=false; i!=-1; i=edges[i].next)
{
if (edges[i].flow&&dis[v=edges[i].to]>(tmp=dis[u]+edges[i].cost))
{
dis[v]=tmp;
p[v]=i^1;
if (vis[v]) continue;
vis[q[r++]=v]=true;
if (r>=maxn) r=0;
}
}
}
return p[dest]>=0;
}
int spfaflow()
{
int i,ret=0,delta;
while (spfa())
{
//按记录原路返回求流量
for (i=p[dest],delta=oo; i>=0; i=p[edges[i].to])
{
delta=min(delta,edges[i^1].flow);
}
for (int i=p[dest]; i>=0; i=p[edges[i].to])
{
edges[i].flow+=delta;
edges[i^1].flow-=delta;
}
ret+=delta*dis[dest];
}
return ret;
}
int main()
{
// freopen("cin.txt","r",stdin);
int n,m;
while(~scanf("%d%d",&n,&m))
{
prepare(2*n*n+3,2*n*n+2,2*n*n+1);
addedge(2*n*n+2,0,m,0);
// addedge(2*n*n,dest,oo,0);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
int x;
scanf("%d",&x);
addedge((i-1)*n+j,(i-1)*n+j+n*n,oo,0);
addedge((i-1)*n+j,(i-1)*n+j+n*n,1,-x);
if(i<n)addedge((i-1)*n+j+n*n,(i)*n+j,oo,0);
if(j<n)addedge((i-1)*n+j+n*n,(i-1)*n+j+1,oo,0);
addedge(0,(i-1)*n+j,oo,0);
addedge((i-1)*n+j+n*n,n*n*2+1,oo,0);
}
}
printf("%d\n",-spfaflow());
}
return 0;
}
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