POJ 3422 Kaka's Matrix Travels 费用流

Kaka's Matrix Travels

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7465		Accepted: 3004

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

POJ Monthly--2007.10.06, Huang, Jinsong

 

题目大意：K取方格数

 

题目分析：拆点，每一个点 i 拆成 i 、i'，由于每一个格子的价值只能取一次，因此建边（i，i'，1，w），又由于每一个格子能屡次通过，因此建边（i，i'，oo，0）。对于其余的点，沿着路径建边便可。

 

代码以下：

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define REP(i, n) for(int i = 1; i <= n; ++i)
#define MS0(X) memset(X,  0, sizeof X)
#define MS1(X) memset(X, -1, sizeof X)
using namespace std;
const int maxE = 3000000;
const int maxN = 5005;
const int maxM = 55;
const int oo = 0x3f3f3f3f;
struct Edge{
    int v, c, w, n;
};
Edge edge[maxE];
int adj[maxN], l;
int d[maxN], cur[maxN], Minflow;
int inq[maxN], Q[maxE], head, tail;
int cost, flow, s, t;
int n, m, nn, a[maxM][maxM];
void addedge(int u, int v, int c, int w){
    edge[l].v = v; edge[l].c = c; edge[l].w =  w; edge[l].n = adj[u]; adj[u] = l++;
    edge[l].v = u; edge[l].c = 0; edge[l].w = -w; edge[l].n = adj[v]; adj[v] = l++;
}
int SPFA(){
    memset(d, oo, sizeof d);
    memset(inq, 0, sizeof inq);
    head = tail = 0;
    d[s] = 0;
    Minflow = oo;
    cur[s] = -1;
    Q[tail++] = s;
    while(head != tail){
        int u =  Q[head++];
        inq[u] = 0;
        for(int i = adj[u]; ~i; i = edge[i].n){
            int v = edge[i].v;
            if(edge[i].c && d[v] > d[u] + edge[i].w){
                d[v] = d[u] + edge[i].w;
                cur[v] = i;
                Minflow = min(edge[i].c, Minflow);
                if(!inq[v]){
                    inq[v] = 1;
                    Q[tail++] = v;
                }
            }
        }
    }
    if(d[t] == oo) return 0;
    flow += Minflow;
    cost += Minflow * d[t];
    for(int i = cur[t]; ~i; i = cur[edge[i ^ 1].v]){
        edge[i].c -= Minflow;
        edge[i ^ 1].c += Minflow;
    }
    return 1;
}
int MCMF(){
    flow = cost = 0;
    while(SPFA());
    return cost;
}
void work(){
    REP(i, n) REP(j, n) scanf("%d", &a[i][j]);
    MS1(adj);
    l = 0;
    nn = n * n;
    s = 0;
    t = (nn << 1) + 1;
    addedge(s, 1, m, 0);
    addedge(nn << 1, t, m, 0);
    REP(i, n) REP(j, n){
        int ij = (i - 1) * n + j;
        addedge(ij, ij + nn, 1, -a[i][j]);
        addedge(ij, ij + nn, oo, 0);
        if(i < n) addedge(ij + nn, i * n + j, oo, 0);
        if(j < n) addedge(ij + nn, (i - 1) * n + j + 1, oo, 0);
    }
    printf("%d\n", -MCMF());
}
int main(){
    while(~scanf("%d%d", &n, &m)) work();
    return 0;
}

POJ 3422