Codeforces 362E Petya and Pipes cost flow diagram

Test instructions

Add capacity to some sides of a network, increasing the sum of the maximum to K, making the maximum flow maximum.

Cost flow: The cost of increasing unit traffic on an edge.

Then it is possible to build 2 edges between 2 points, and the first given edge (u,v,x,0) has an edge cost of 0

At the same time the other side (u,v,k,1) costs 1, you can find the maximum flow by limiting the corresponding cost at the time of augmentation

Individuals feel that the reason for this is that each time the credit is optimal. Therefore, the optimal solution can be obtained by limiting the final cost not exceeding K

#include <map>#include<Set>#include<list>#include<cmath>#include<ctime>#include<deque>#include<stack>#include<queue>#include<cctype>#include<cstdio>#include<string>#include<vector>#include<climits>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#defineLL Long Long#definePI 3.1415926535897932626using namespacestd;intgcdintAintb) {returnA% b = =0? B:GCD (b, a%b);}Const intMAXN = the;Const intINF =0x3f3f3f3f;structnode{intU,v,next; intFlow,cap,cost;} EDGE[MAXN* MAXN *4];intCnt,src,tag;intc,f;intK,n;queue<int>Q;BOOLINQ[MAXN];intD[MAXN];intHEAD[MAXN],P[MAXN];inttot =0;voidinit () {memset (head,-1,sizeof(head)); Tot=0;}voidAdd_edge (intUintVintCapintCost ) {edge[cnt].u=u; EDGE[CNT].V=v; Edge[cnt].cap=cap; Edge[cnt].flow=0; Edge[cnt].cost=Cost ; Edge[cnt].next=Head[u]; Head[u]= cnt++; //ReverseEDGE[CNT].V =u; EDGE[CNT].U=v; Edge[cnt].flow=0; Edge[cnt].cap=0; Edge[cnt].cost= -Cost ; Edge[cnt].next=Head[v]; HEAD[V]= cnt++;}BOOLSPFA (intSintt) {     while(!q.empty ()) Q.pop (); memset (INQ,false,sizeof(INQ)); memset (d,0x3f,sizeof(d)); Memset (P,-1,sizeof(p)); D[s]=0;    Q.push (s); Inq[s]=true;  while(!Q.empty ()) {        intU =Q.front (); Q.pop (); Inq[u]=false;  for(inti = Head[u]; I! =-1; i =Edge[i].next) {            intv =edge[i].v; if(D[v] > D[u] + edge[i].cost && edge[i].cap >Edge[i].flow) {D[v]= D[u] +Edge[i].cost; P[V]=i; if(!Inq[v])                    {Q.push (v); INQ[V]=true; }            }        }    }    if(D[tag] = = INF)return false; intA =INF;  for(inti = P[tag]; I! =-1; i =p[edge[i].u]) a= Min (A,edge[i].cap-Edge[i].flow); if(C + d[tag] * a >K) {F+ = (k-c)/D[tag]; return false; }    return true;}voidSlove () {C= F =0;  while(SPFA (Src,tag)) {intA =INF;  for(inti = P[tag]; I! =-1; i =p[edge[i].u]) a= Min (A,edge[i].cap-Edge[i].flow);  for(inti = P[tag]; I! =-1; i =P[EDGE[I].U]) {Edge[i].flow+=A; Edge[i^1].flow-=A; } C+ = D[tag] *A; F+=A; }}intMain () { while(SCANF ("%d%d", &n,&k)! =EOF)        {init ();  for(inti =1; I <= N; i++)             for(intj =1; J <= N; J + +)        {            intx; scanf ("%d",&x); if(x) {Add_edge (i,j,x,0); Add_edge (I,j,k,1); }} src=1; Tag=N;        Slove (); printf ("%d\n", F); }    return 0;}

Codeforces 362E Petya and Pipes cost flow diagram