深海机器人问题

题目描述

题解：

最大费用最大流。

建图很简单，就是将机器人作为流，进入就从$S$向内流，出来就从图向$T$流。

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 300
#define ll long long
const int inf = 0x3f3f3f3f;
const ll Inf  = 0x3f3f3f3f3f3f3f3fll;
inline int rd()
{
    int f=1,c=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
    return f*c;
}
int a,b,P,Q,S,T,hed[N],cnt=-1;
int _id(int x,int y){return x*(Q+1)+y;}
struct EG
{
    int to,nxt;
    ll w,c;
}e[10*N];
void ae(int f,int t,ll w,ll c)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    e[cnt].w = w;
    e[cnt].c = c;
    hed[f] = cnt;
}
queue<int>q;
ll dis[N],fl[N];
int pre[N],fa[N];
bool vis[N];
bool spfa()
{
    memset(dis,0x3f,sizeof(dis));
    dis[S] = 0,fl[S] = Inf,vis[S] = 1;
    q.push(S);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int j=hed[u];~j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(e[j].w&&dis[to]>dis[u]+e[j].c)
            {
                dis[to] = dis[u]+e[j].c;
                fl[to] = min(fl[u],e[j].w);
                pre[to] = j,fa[to] = u;
                if(!vis[to])
                {
                    vis[to] = 1;
                    q.push(to);
                }
            }
        }
        vis[u] = 0;
    }
    return dis[T]!=Inf;
}
ll mcmf()
{
    ll ret = 0;
    while(spfa())
    {
        ret+=fl[T]*dis[T];
        int u = T;
        while(u!=S)
        {
            e[pre[u]].w-=fl[T];
            e[pre[u]^1].w+=fl[T];
            u=fa[u];
        }
    }
    return ret;
}
int main()
{
    a = rd(),b = rd(),P = rd(),Q = rd();
    memset(hed,-1,sizeof(hed));
    for(int i=0;i<=P;i++)
        for(int c,j=0;j<Q;j++)
        {
            int f = _id(i,j);
            int t = _id(i,j+1);
            c = rd();
            ae(f,t,1,-c);
            ae(t,f,0,c);
            ae(f,t,Inf,0);
            ae(t,f,0,0);
        }
    for(int i=0;i<=Q;i++)
        for(int c,j=0;j<P;j++)
        {
            int f = _id(j,i);
            int t = _id(j+1,i);
            c = rd();
            ae(f,t,1,-c);
            ae(t,f,0,c);
            ae(f,t,Inf,0);
            ae(t,f,0,0); 
        }
    S = _id(P,Q)+1,T = S+1;
    for(int w,x,y,i=1;i<=a;i++)
    {
        w = rd(),x = rd(),y = rd();
        int t = _id(x,y);
        ae(S,t,w,0);
        ae(t,S,0,0);
    }
    for(int w,x,y,i=1;i<=b;i++)
    {
        w = rd(),x = rd(),y = rd();
        int f = _id(x,y);
        ae(f,T,w,0);
        ae(T,f,0,0);
    }
    printf("%lld
",-mcmf());
    return 0;
}