POJ 2516 Minimum cost or minimum charge flow

Test instructions

There are k items, n stores and M stations, each store has a demand for each commodity shopneed[n][k], each depot has a stock of each commodity supply[m][k], for the category K items, he from the supply station J to store I cost Cost[k][i][j], Ask each store to meet the minimum cost of demand for each item.

Analysis:

The typical minimum cost maximum flow problem, but because each commodity's demand and the inventory quantity are very small (<=3), each point splits the new point also to be few, therefore may use the +km method to do.

Code:

POJ 2516//sep9#include <iostream>using namespace Std;const int Maxn=220;int w[maxn][maxn];int mapX[maxN],mapY[ Maxn];int lx[maxn],ly[maxn],linky[maxn];int visx[maxn],visy[maxn];int slack[maxn];int nx,ny,n,m,k;int ShopNeed[maxN ][maxn];int supply[maxn][maxn];int cost[maxn][maxn][maxn];bool Find (int x) {visx[x]=true;for (int y=1;y<=ny;++y) { if (Visy[y]) continue;int t=lx[x]+ly[y]-w[x][y];if (t==0) {visy[y]=true;if (linky[y]==-1| | Find (Linky[y])) {Linky[y]=x;return true;}} else if (slack[y]>t) slack[y]=t;} return false;} int KM () {int I,j;memset (linky,-1,sizeof (Linky)), memset (ly,0,sizeof (ly)); for (i=1;i<=nx;++i) for (J=1,lx[i]=int_ MIN;J&LT;=NY;++J) if (W[i][j]>lx[i]) lx[i]=w[i][j];for (int x=1;x<=nx;++x) {for (i=1;i<=ny;++i) slack[i]=INT_ Max;while (1) {memset (visx,0,sizeof (VISX)); Memset (Visy,0,sizeof (Visy)); if (find (x)) Break;int d=int_max;for (i=1;i <=ny;++i) if (!visy[i]&&slack[i]<d) d=slack[i]; if (D==int_max) return 1;for (I=1;i<=nx;++i) if (Visx[i]) lx[i]-=d;for (i=1;i<=ny;++i) if (Visy[i]) Ly[i]+=d;elseslack[i]-=d;}} int Result=0;for (i=1;i<=ny;++i) if (linky[i]>-1) Result+=w[linky[i]][i];return result; int main () {while (scanf ("%d%d%d", &n,&m,&k) ==3) {if (n==0&&m==0&&k==0) Break;int i,j,t; for (I=1;i<=n;++i) for (t=1;t<=k;++t) scanf ("%d", &shopneed[i][t]); for (j=1;j<=m;++j) for (t=1;t<=k;++ T) scanf ("%d", &supply[j][t]); for (t=1;t<=k;++t) for (i=1;i<=n;++i) for (j=1;j<=m;++j) scanf ("%d",& COST[T][I][J]); int ans=0,p,q;for (t=1;t<=k;++t) {nx=0;ny=0;for (i=1;i<=n;++i) for (p=1;p<=shopneed[i][t];++ p) mapx[++nx]=i;for (J=1;J&LT;=M;++J) for (q=1;q<=supply[j][t];++q) mapy[++ny]=j;for (i=1;i<=nx;++i) for (j=1;j &LT;=NY;++J) W[i][j]=-cost[t][mapx[i]][mapy[j]];int x=km (); if (x>0) {ans=-1;break;} Elseans-=x;} printf ("%d\n", ans);}  return 0;}

POJ 2516 Minimum cost or minimum charge flow