題目大意

在一個平面直角座標系中，機器人只能往右和上採集標本，每個格點都有不同的價值，現在若干個機器人從某點出發目的地為某點，問採集到的最大價值

分析

其實這道題類比於K取方格數，容易建出這樣一張圖

然後跑一遍最大費用最大流就可以了，但是我把費用取反，跑的是最小費用最大流

程式碼

#include <cstdio>
#include <deque>
#include <cstring>
#define rr register
#define id(x,y) ((x-1)*m+y)
using namespace std;
 

const int inf=707406378;
struct node{
    int y,w,f,next;
}e[3001];
int ss,tt,n,m,s,t,ans,k=1,ls[301],dis[301]; bool v[301];
inline void add(int x,int y,int w,int f){
    e[++k]=(node){y,w,f,ls[x]}; ls[x]=k;
    e[++k]=(node){x,0,-f,ls[y]}; ls[y]=k;
}
inline signed spfa(){
    memset(v,0,sizeof(v)); memset(
 
dis,127/3,sizeof(dis));
    dis[t]=0; v[t]=1; rr deque<int>q; q.push_back(t);
    while (q.size()){
        rr int x=q.front(); q.pop_front();
        for (rr int i=ls[x];i;i=e[i].next)
        if (e[i^1].w&&dis[e[i].y]>dis[x]-e[i].f){
            dis[e[i].y]=dis[x]-e[i].f;
            if
 
 (!v[e[i].y]){
                v[e[i].y]=1;
                if (q.size()&&dis[e[i].y]<dis[q.front()])
                q.push_front(e[i].y); else q.push_back(e[i].y);
            }
        }
        v[x]=0;
    }
    return dis[s]<707406378;
}
inline signed dfs(int x,int now){
    if (x==t) {v[t]=1; return now;}
    rr int rest=0,f; v[x]=1;
    for (rr int i=ls[x];i;i=e[i].next)
    if (!v[e[i].y]&&e[i].w&&dis[e[i].y]+e[i].f==dis[x]){
        rest+=(f=dfs(e[i].y,min(e[i].w,now-rest)));
        if (f) ans-=f*e[i].f,e[i].w-=f,e[i^1].w+=f;
        if (rest==now) break;
    }
    return rest;
}
inline void answ(){
    while (spfa()){
        v[t]=1;
        while (v[t]){
            memset(v,0,sizeof(v));
            dfs(s,1e9);
        }
    }
}
signed main(){
    scanf("%d%d%d%d",&ss,&tt,&n,&m); s=++n*++m+1,t=s+1;
    for (rr int i=1;i<=n;++i)
    for (rr int j=1,x;j<m;++j){
        scanf("%d",&x);
        add(id(i,j),id(i,j+1),1,-x);
        add(id(i,j),id(i,j+1),inf,0);
    }
    for (rr int j=1;j<=m;++j)
    for (rr int i=1,x;i<n;++i){
        scanf("%d",&x);
        add(id(i,j),id(i+1,j),1,-x);
        add(id(i,j),id(i+1,j),inf,0);		
    }
    for (rr int i=1,x,y,w;i<=ss;++i){
    	scanf("%d%d%d",&w,&x,&y);
    	add(s,id(++x,++y),w,0);
    }
    for (rr int i=1,x,y,w;i<=tt;++i){
    	scanf("%d%d%d",&w,&x,&y);
    	add(id(++x,++y),t,w,0);
    }
    answ();
    printf("%d",ans);
    return 0;
}