网络流二十四题之运输问题

这道题真的很水，如果和前面的几道题比一下的话。
建模很轻松，按照题意无脑建模就行：
1。源点向仓库连一条边，容量为该仓库的货物，费用为0，代表该仓库能提供多少货物
2.每个仓库向每个商店连一条边，容量INF（其实a[i]也可以），费用为Cij
3.每个仓库向汇点连一条边，容量为bi，费用0，代表这个商店接受正好bi的货物。
由于数据sum(a) == sum(b) 就不用考虑别的问题，只需要跑两次费用流，第二次的时候将费用取反就行。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int MAXN = 100000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int from,to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
    memset(edge,0,sizeof(edge));
}
void addedge(int u,int v,int cap,int cost)
{
    edge[tol].from = u;
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].from = v;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i = 0; i < N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] + edge[i].cost )
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)
        return false;
    else
        return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}
void print_all()
{
    for(int i = 0; i<tol; i+=2)
    {
        printf("%d->%d cap = %d flow  = %d cost = %d\n",edge[i].from,edge[i].to,edge[i].cap,edge[i].flow,edge[i].cost);
    }
}
int a[1001],b[1001];
int c[1001][1001];
int main()
{
    int n,m;
    scanf("%d%d",&m,&n);
    for(int i  =1;i<=m;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int j = 1;j<=n;j++)
    {
        scanf("%d",&b[j]);
    }
    for(int i = 1;i<=m;i++)
    {
        for(int j = 1;j<=n;j++)
        {
            scanf("%d",&c[i][j]);
        }
    }
    init(n + m + 2);
    for(int i = 1;i<=m;i++)
    {
        addedge(0,i,a[i],0);
    }
    for(int i = 1;i<=n;i++)
    {
        addedge(i + m ,n+m+1,b[i],0);
    }
    for(int i = 1;i<=m;i++)
    {
        for(int j = 1;j<=n;j++)
        {
            addedge(i,j +m,a[i],c[i][j]);
        }
    }
    int cost;
    int ans = minCostMaxflow(0,n+m + 1,cost);
    printf("%d\n",cost);
     init(n + m + 2);
    for(int i = 1;i<=m;i++)
    {
        addedge(0,i,a[i],0);
    }
    for(int i = 1;i<=n;i++)
    {
        addedge(i + m ,n+m+1,b[i],0);
    }
    for(int i = 1;i<=m;i++)
    {
        for(int j = 1;j<=n;j++)
        {
            addedge(i,j +m,a[i],-c[i][j]);
        }
    }
    ans = minCostMaxflow(0,n+m+1,cost);
    printf("%d\n",-cost);
    return 0;
}