POJ3422 Kaka's Matrix Travels 【费用流】*

POJ3422 Kaka’s Matrix Travels

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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

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题意大概就是给你一个n*n的矩形，每一个方格中有一个权值，定义一次行走是从左上角走到右下角，而且每一次移动只能向右边或者下边移动，现在我们要行走k次，并使经过的点集的和最大（每个点值计算一次）

我们很容易想到把每个点的权值作为这个点某种意义上的费用（因为只能计算一次，后面会介绍建图方法），然后我们想最大化这个费用，所以对于这种问题我们可以见图的时候将边权取相反数，然后再跑出最小费用大流的时候将费用取相反数，不难发现这个时候的费用是最大化的

然后我们怎么处理一个节点的边权只能计算一次这个条件呢？

不妨这样思考，我们把一个点i拆成两个点i" role="presentation">ii和i+n∗n" role="presentation">i+n∗ni+n∗n，在i" role="presentation">ii和i+n∗n" role="presentation">i+n∗ni+n∗n之间连接两条弧：

1：一条容量为1，费用为负边权

2：一条容量为k-1，费用为0

我们可以很容易得到如果有流量经过弧2，一定有流量经过弧1，不然就不满足最小费用的性质

又对于两个节点i" role="presentation">ii和j" role="presentation">jj，满足j在i的下边或右边，即i" role="presentation">ii可以到达j" role="presentation">jj，那么我们可以把i+n∗n" role="presentation">i+n∗ni+n∗n连接到j" role="presentation">jj，这样流经i" role="presentation">ii的流量就可以流到j" role="presentation">jj了

现在我们只需要保证源点S流到第一个节点（坐标(1,1)）的容量是k就好了

整理一下构图思路：

1 从S" role="presentation">SS连接到第一个节点(1,1)" role="presentation">(1,1)(1,1)，容量为k

2 从i" role="presentation">ii到i+n∗n" role="presentation">i+n∗ni+n∗n连接两条弧：

     " role="presentation">          1：一条容量为1，费用为负边权

     " role="presentation">          2：一条容量为k-1，费用为0

3 对于i" role="presentation">ii到可以到达的j" role="presentation">jj连接一条从i+n∗n" role="presentation">i+n∗ni+n∗n到j" role="presentation">jj的边

4 从最后一个节点(n,n)" role="presentation">(n,n)(n,n)连接到汇点T" role="presentation">TT

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define N 100010
#define INF 0x3f3f3f3f
struct Edge{
    int u,v,cap,flow,cost;
    Edge(int xu,int xv,int xcap,int xflow,int xcost){
        u=xu;v=xv;cap=xcap;flow=xflow;cost=xcost;
    }
};
struct MCMF{
    int s,t,f[N],p[N],d[N];
    bool inq[N];
    vector<Edge> E;
    vector<int> G[N];
    void add(int u,int v,int cap,int cost){
        E.push_back(Edge(u,v,cap,0,cost));
        E.push_back(Edge(v,u,0,0,-cost));
        int m=E.size();
        G[u].push_back(m-2);
        G[v].push_back(m-1);
    }
    bool SPFA(int &flow,int &cost){
        memset(inq,0,sizeof(inq));
        memset(d,0x3f,sizeof(d));
        queue<int> Q;Q.push(s);
        d[s]=0;f[s]=INF;
        while(!Q.empty()){
            int u=Q.front();Q.pop();
            inq[u]=0;
            for(int i=0;i<G[u].size();i++){
                Edge e=E[G[u][i]];
                if(d[e.v]>d[u]+e.cost&&e.cap>e.flow){
                    d[e.v]=d[u]+e.cost;
                    p[e.v]=G[u][i];
                    f[e.v]=min(f[u],e.cap-e.flow);
                    if(!inq[e.v]){
                        Q.push(e.v);
                        inq[e.v]=1;
                    }
                }
            }
        }
        if(d[t]==INF)return false;
        flow+=f[t];cost+=f[t]*d[t];
        int u=t;
        while(u!=s){
            E[p[u]].flow+=f[t];
            E[p[u]^1].flow-=f[t];
            u=E[p[u]].u;
        }
        return true;
    }
    int Min_cost_Max_flow(){
        int flow=0,cost=0;
        while(SPFA(flow,cost));
        return cost;
    }
}mcmf;
int n,k;
int g[100][100];
int Index(int x,int y){return (x-1)*n+y;}
int main(){
    scanf("%d%d",&n,&k);
    mcmf.s=0;mcmf.t=2*n*n+1;
    mcmf.add(0,1,k,0);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&g[i][j]);
            int tmp=Index(i,j);
            mcmf.add(tmp,tmp+n*n,1,-g[i][j]);
            mcmf.add(tmp,tmp+n*n,k-1,0);
            if(i<n){
                int tmp1=Index(i+1,j);
                mcmf.add(tmp+n*n,tmp1,INF,0);
            }
            if(j<n){
                int tmp1=Index(i,j+1);
                mcmf.add(tmp+n*n,tmp1,INF,0);
            }
        }
    }
    mcmf.add(2*n*n,2*n*n+1,k,0);
    printf("%d",-mcmf.Min_cost_Max_flow());
    return 0;
}