POJ 2516 Minimum Cost(最小费用最大流)

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
const int maxn = 110;
const int maxm = 10000;
const int inf = 1e8;
#define MIN INT_MIN
#define MAX 1e6
#define LL long long
#define init(a) memset(a,0,sizeof(a))
#define FOR(i,a,b) for(int i = a;i<b;i++)
#define max(a,b) (a>b)?(a):(b)
#define min(a,b) (a>b)?(b):(a)
using namespace std;
struct node
{
int u,v,w,cap,next;
}edge[maxm];
int pre[maxn],dis[maxn],head[maxn],cnt;
bool vis[maxn];
int n,m;
void add(int u,int v,int c,int cap)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].w=c;
edge[cnt].cap=cap;
edge[cnt].next=head[u];
head[u]=cnt++;

edge[cnt].u=v;
edge[cnt].v=u;
edge[cnt].w=-c;
edge[cnt].cap=0;
edge[cnt].next=head[v];
head[v]=cnt++;
}
int spfa(int s,int t)
{
queue<int>q;
while(q.empty()==false) q.pop();
q.push(s);
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
FOR(i,s,t+1)
dis[i] = inf;
dis[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u] = 0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
if(edge[i].cap && dis[edge[i].v]>dis[u]+edge[i].w)
{
dis[edge[i].v]=dis[u]+edge[i].w;
pre[edge[i].v] = i;
if(!vis[edge[i].v])
{
vis[edge[i].v]=1;
q.push(edge[i].v);
}
}
}

}
if(dis[t] != inf)
return 1;
else
return 0;
}
int MinCostMaxFlow(int s,int t)
{
int flow=0,cost=0;
while(spfa(s,t))
{
int df = inf;
for(int i = pre[t];i!=-1;i=pre[edge[i].u])
{
if(edge[i].cap<df)
df = edge[i].cap;
}
flow += df;
for(int i=pre[t];i!=-1;i=pre[edge[i].u])
{
edge[i].cap -= df;
edge[i^1].cap += df;
}
cost += dis[t] * df;
}
return cost;
}
void initt()
{
cnt=0;
memset(head,-1,sizeof(head));
}

int main()
{
int gong[51],qiu[51];
int k,s,t;
int need[200][200],gei[200][200];
while(scanf("%d%d%d",&n,&m,&k),n,m,k)
{
init(need);init(gei);
init(gong);init(qiu);
s = 0,t = n+m+1;
FOR(i,1,n+1)
{
FOR(j,1,k+1)
{
scanf("%d",&need[i][j]);
qiu[j] += need[i][j];
}
}
FOR(i,1,m+1)
{
FOR(j,1,k+1)
{
scanf("%d",&gei[i][j]);
gong[j] += gei[i][j];
}
}
bool flag = 1;
FOR(i,1,k+1)
{
if(gong[i]<qiu[i])//如果单种商品的供给量不满足该商品的需求量，直接打印-1
{
flag = 0;
break;
}
}
int cost,yao,money = 0;

FOR(ji,1,k+1)
{
initt();

FOR(i,1,n+1)
{
FOR(j,1,m+1)
{
scanf("%d",&cost);
if(flag==0) continue;
add(j,m+i,cost,inf);//供给商和店主的流量为正无穷
}
}
if(flag==0) continue;
FOR(i,1,n+1)
{
add(m+i,t,0,need[i][ji]);//店铺和汇点的费用为0，流量为店对该种商品的需求
}
FOR(i,1,m+1)
{
add(s,i,0,gei[i][ji]);//源点对供应商的费用为0，流量为供货商对该种商品的供给
}
money += MinCostMaxFlow(s,t);
}
(flag==0)?puts("-1"):printf("%d\n",money);
}
return 0;
}