网络流二十四题深海机器人问题

其实仔细想想，这道题和一开始的一道问题很想像，这道题都是边权，就不用拆点了
1.源点像每个可以发个机器人的地方连一条边，容量为可以放的个数，所有回收点向汇点连一条边，容量是可以返回的个数。费用都是0；
2.然后按照方向连点，容量1，费用为该边权值，代表这条边的资源只能采集一次。但是还需要一条容量INF，费用0的边，代表这条边可以通过，但是不一定要采集资源。
之后跑一个最大费用流

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#define ss 9999
#define tt 10000
using namespace std;
const int MAXN = 100000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int 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));
}
void addedge(int u,int v,int cap,int cost)
{
    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].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;
}
int a,b,p,q;
int rid(int x,int y) {return x*(q + 1) + y;}
int main()
{
    scanf("%d%d%d%d",&a,&b,&p,&q);
    init(tt+1);
    for(int i = 1;i<=p+1;i++)
    {
        for(int j = 1;j<=q;j++)
        {
            int x;
            scanf("%d",&x);
            addedge(rid(i-1,j-1),rid(i-1,j),1,-x);
            addedge(rid(i-1,j-1),rid(i-1,j),INF,0);
        }
    }
    for(int i = 1;i<=q+1;i++)
    {
        for(int j = 1;j<=p;j++)
        {
            int x;
            scanf("%d",&x);
            addedge(rid(j-1,i-1),rid(j,i-1),1,-x);
            addedge(rid(j-1,i-1),rid(j,i-1),INF,0);
        }
    }
    for(int i = 0;i<a;i++)
    {
        int k,x,y;
        scanf("%d%d%d",&k,&x,&y);
        addedge(ss,rid(x,y),k,0);
    }
    for(int j = 0;j<b;j++)
    {
        int k,x,y;
        scanf("%d%d%d",&k,&x,&y);
        addedge(rid(x,y),tt,k,0);
    }
    int cost = 0;
    minCostMaxflow(ss,tt,cost);
    printf("%d",-cost);
    return 0;
}