POJ 2526 Minimum Cost "minimum charge maximum Flow"

Title Link: http://poj.org/problem?id=2516

Test instructions
N Shop owner m suppliers k kinds of goods to the owner of the K-type goods and the supplier of K-type cargo stock and K type of transportation costs.

Solution: The K-cost flow, respectively, the cost of each kind of goods. Source point to the supply side, the shopkeeper to the meeting point to build the edge, the cost is 0, the capacity of 1. Then supply to the shopkeeper to build the side, the cost of the matrix, the capacity is infinite.

Code:

/*poj 2195 going home adjacency matrix form minimum cost maximum flow * /#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>#include <queue>using namespace STD;//************************************************************//Minimum cost maximum flow algorithm//SPFA to find the shortest way//adjacency matrix form//Initialize: Cap: Capacity, no edge for 0//cost: Cost, symmetrical form, 0 without side//c is the minimum cost//f is the maximum flow//*******************************************************Const intMAXN = -;Const intINF =0x3fffffff;intCAP[MAXN][MAXN];//capacity, No side for 0intFLOW[MAXN][MAXN];//cost matrix is symmetrical, there is a charge of I to J, then the cost of J to I is the opposite numberintCOST[MAXN][MAXN];//CostintN///vertex number 0~n-1intF//Maximum flowintC//Minimum costintStart, End;//Source point and meeting pointBOOLVIS[MAXN];//In the queue flagintQUE[MAXN];intPRE[MAXN];intDIST[MAXN];//s-t path Minimum costBOOLSPFA () {intFront =0, rear =0; for(intU =0; U <= N; u++) {if(U = = start)            {que[rear++] = u; Dist[u] =0; Vis[u] =true; }Else{Dist[u] = INF; Vis[u] =false; }    } while(Front! = rear) {intU = que[front++]; Vis[u] =false;if(Front >= maxn) front =0; for(intv =0; V <= N; v++) {if(Cap[u][v]>flow[u][v] && Dist[v]>dist[u] + cost[u][v])                {Dist[v] = Dist[u] + cost[u][v]; PRE[V] = u;if(!vis[v]) {Vis[v] =true; que[rear++] = v;if(Rear >= MAXN) rear =0; }            }        }    }if(Dist[end] >= INF)return false;return true;}voidMincostmaxflow () {memset(Flow,0,sizeof(flow)); c = f =0; while(SPFA ()) {intMin = INF; for(intu = End; U! = start; u = pre[u]) min = min (min, cap[pre[u]][u]-flow[pre[u]][u]); for(intu = End; U! = start;            u = pre[u]) {Flow[pre[u]][u] + = Min;        Flow[u][pre[u]] = Min;        } c + = dist[end] * Min;    f + = Min; }}//************************************************************inttmpinta[10000][ -];intb[10000][ -];intMain () {intN, M, K; while(~scanf("%d%d%d", &n, &m, &k)) {if(N = =0&& M = =0&& K = =0) Break;memsetA0,sizeof(a));memset(b,0,sizeof(b)); Start =0; n = n + m+2; End = M + N +1;intNeed =0; for(inti =1; I <= N; i++) { for(intK =1; K <= k;k++) {scanf("%d", &a[i][k]);            Need + = A[i][k]; }        } for(inti =1; I <= M; i++) { for(intK =1; K <= K; k++) {scanf("%d", &b[i][k]); }        }intAns =0;intres =0; for(intKK =1; KK <= K; kk++) {memset(Cap,0,sizeof(CAP));memset(Cost,0,sizeof(cost)); for(inti =1; I <= M; i++)//source points to the supply point to build edgesCap[start][i] = B[i][kk]; for(inti =1; I <= N; i++)//Shopkeeper to the meeting point to build the sideCap[m + i][end] = A[i][kk]; for(inti =1; I <= N; i++) for(intj =1; J <= M; J + +) {scanf("%d", &tmp);                    Cost[j][i + M] = tmp;                    Cost[i + m][j] =-tmp; Cap[j][i + M] =1000000;            } mincostmaxflow ();            Ans + = c;        Res + = f; }if(res = = need)printf("%d\n", ans);Else            printf(" -1\n"); }return 0;}

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Poj 2526 Minimum cost min. Max Flow