Poj 3422 Kaka's matrix travels (billing flow)

KaKa's matrix travels

Time limit:1000 ms		Memory limit:65536 K
Total submissions:7743		Accepted:3111

Description

OnN×NChessboard with a non-negative number in each grid, Kaka starts his matrix travelsSum= 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 numberSumIn each grid the rook visited, and replaces it with zero. It is not difficult to know the maximumSumKaKa can obtain for his first travel. Now Kaka is wondering what is the maximumSumHe can obtain after hisKTh travel. NoteSumIs accumulative duringKTravels.

Input

The first line contains two integersNAndK(1 ≤N≤ 50, 0 ≤K≤ 10) described above. The followingNLines represents the Matrix. You can assume the numbers in the matrix are no more than 1000.

Output

The maximumSumKaKa can obtain after hisKTh travel.

Sample Input

 
3 21 2 30 2 11 4 2
 

Sample output

 
15
 

I can't speak out. Open the array bigger !!

This question needs to be split when creating a graph. Each grid is split into two vertices, one being the output vertex u, and the other being the entry vertex V, and the other being the U-> v.

One capacity is 1, the weight is the amount of money in the grid; the other capacity is K-1, the weight is 0. Because once the money is collected, there is no money.

There is a path from V to the next grid (right and bottom). The capacity is K and the fee is zero.

The super source is 0, and the super sink is N * n * 2 + 1.

# Include "stdio. H "# include" string. H "# include" queue "using namespace STD; # define n 50000 const int INF = 1 <20; struct node {int U, V, C, F, next ;} E [N * 2]; int pre [N], DIS [N], vis [N], head [N], T; void Add1 (int u, int V, int C, int f) {e [T]. U = u; E [T]. V = V; E [T]. C = C; E [T]. F = f; E [T]. next = head [u]; head [u] = T ++;} void add (int u, int V, int C, int f) {Add1 (u, v, c, F); Add1 (v, U,-C, 0);} int spfa (int s, int N) {int I, U, V; for (I = s; I <= N; I + +) Dis [I] = inf; DIS [s] = 0; memset (PRE,-1, sizeof (pre); memset (VIS, 0, sizeof (VIS); queue <int> q; q. push (s); While (! Q. Empty () {u = Q. Front (); q. Pop (); vis [u] = 0; for (I = head [u]; I! =-1; I = E [I]. next) {v = E [I]. v; If (E [I]. F & dis [v]> dis [u] + E [I]. c) {dis [v] = dis [u] + E [I]. c; Pre [v] = I; If (! Vis [v]) {vis [v] = 1; q. Push (v) ;}}} if (pre [N]! =-1) return 1; return 0;} void solve (int s, int N) {int I, j, cost, minf; cost = 0; while (spfa (S, n) {minf = inf; for (I = pre [N]; I! =-1; I = pre [E [I]. u]) {If (minf> E [I]. f) minf = E [I]. f ;}for (I = pre [N]; I! =-1; I = pre [E [I]. u]) {J = I ^ 1; E [I]. f-= minf; E [J]. F + = minf;} cost + = minf * Dis [N];} printf ("% d \ n",-cost);} int main () {int I, j, N, K, U, C; while (scanf ("% d", & N, & K )! =-1) {T = 0; memset (Head,-1, sizeof (head); for (I = 1; I <= N; I ++) {for (j = 1; j <= N; j ++) {scanf ("% d", & C); U = (I-1) * n + J; add (u, u + N * n,-C, 1); add (u, u + N * n, 0, k-1); if (I + 1 <= N) add (U + N * n, U + N, 0, k); If (J + 1 <= N) add (U + N * n, U + 1, 0, k) ;}} add (, 0, k); add (2 * n, 2 * n +, K ); solve (* n + 1);} return 0 ;}