Bzoj 1834 ZJOI2010 Network expansion Dinic+ek fee flow

Topic: Given an n-point m-edge of the graph, each edge has a capacity to increase the cost of C, representing the cost of 1 per capacity expansion, the maximum flow and the maximum flow to expand the minimum cost of K

First ask direct run Max Flow

The second question is to connect the starting point of each edge to the end of a traffic is positive infinity, the cost of C of the edge and then the N to the meeting point of a traffic for the ans+k cost of 0 of the edge run the minimum cost maximum flow can be

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 5010# Define INF 0x3f3f3f3f#define S 1#define T (n+1) using namespace std;struct edge{int x,y,f,c;} Edges[m];struct abcd{int To,f,c,next;} Table[m<<2];int head[m],tot=1;int n,m,k,ans,anscost;int dpt[m];void Add (int x,int y,int f,int c) {table[++tot].to =y;table[tot].f=f;table[tot].c=c;table[tot].next=head[x];head[x]=tot;} BOOL BFS () {static int q[m],r,h;int i;memset (dpt,-1,sizeof DPT); R=h=0;q[++r]=s;dpt[s]=1;while (r!=h) {int x=q[++h];for ( I=head[x];i;i=table[i].next) if (table[i].f&&!~dpt[table[i].to]) {dpt[table[i].to]=dpt[x]+1;q[++r]=table[ I].to;if (table[i].to==t) return true;}} return false;} int dinic (int x,int flow) {int i,left=flow;if (x==t) return flow;for (I=head[x];i&&left;i=table[i].next) if ( table[i].f&&dpt[table[i].to]==dpt[x]+1) {int temp=dinic (table[i].to,min (LEFT,TABLE[I].F)), if (!temp) dpt[ Table[i].to]=-1;left-=temp;table[i].f-=temp;table[i^1].f+=temp;} ReturnFlow-left;} BOOL EK () {static int q[65540],flow[m],cost[m],from[m];static bool V[m];static unsigned short r,h;int i;memset (cost,0x3f , sizeof cost); Cost[s]=0;flow[s]=inf;q[++r]=s;while (r!=h) {int x=q[++h];v[x]=0;for (i=head[x];i;i=table[i].next) if ( TABLE[I].F) if (cost[table[i].to]>cost[x]+table[i].c) {cost[table[i].to]=cost[x]+table[i].c;flow[table[i].to]= Min (FLOW[X],TABLE[I].F); From[table[i].to]=i;if (!v[table[i].to]) v[table[i].to]=1,q[++r]=table[i].to;} if (Cost[t]==inf) return false;anscost+=flow[t]*cost[t];for (I=from[t];i;i=from[table[i^1].to]) Table[i].f-=flow[T] , Table[i^1].f+=flow[t];return true;} int main () {int i;cin>>n>>m>>k;for (i=1;i<=m;i++) {scanf ("%d%d", &edges[i].x,&edges[i]. y); scanf ("%d%d", &edges[i].f,&edges[i].c); ADD (edges[i].x,edges[i].y,edges[i].f,0); ADD (edges[i].y,edges[i].x,0,0);} ADD (n,t,inf,0); Add (t,n,0,0), while (BFS ()) Ans+=dinic (S,inf), Table[tot-1].f=k;for (i=1;i<=m;i++) {ADD (edges[i].x,edges[i].y,inf , EDGES[I].C); ADD (edges[i].y,edges[i].x,0,-EDGES[I].C);} while (EK ());cout<<ans<< "&LT;&LT;ANSCOST&LT;&LT;ENDL;}

Bzoj 1834 ZJOI2010 Network expansion Dinic+ek fee flow