P4012 深海機器人問題

題目連結： P4012

題意

在一個網格中給你一些機器人，要求只能向上走，或向左走，每一條邊都有一個寶藏，每個寶藏只能被拿一次。機器人只能在一些指定點開始放，在一些指定點收回來。問最多可以獲得多少財富？

思路

我一開始想的也是費用流，不過那個只能走一次的問題，我一直想不好。

其實只需要再建一條邊就好了，除了add_edge(u,v,1,-val) 再建一條邊add_edge(u,v,INF,0)即可，給圖一個可流的區間。

程式碼

#include <iomanip>
#include <cstring>
#include <cstdlib>
 

#include <cctype>
#include <cstdio>
#include <string>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <sstream>
#include <memory>
 

#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i,j,k) for(int i = (int)j;i <= (int)k;i ++)
#define per(i,j,k) for(int i = (int)j;i >= (int)k;i --)
#define debug(x) cerr<<#x<<" = "<<(x)<<endl
#define mmm(a,b) memset(a,b,sizeof(a))
#define
 
 pb push_back

typedef double db;
typedef long long ll;
const int MAXN = (int)1e3+7;
const int INF = (int)0x3f3f3f3f;

struct edge{
    int to,cap,cost,rev;
    edge(int to = 0,int cap = 0,int cost = 0,int rev = 0):to(to),cap(cap),cost(cost),rev(rev){}
};

int N,M;
vector<edge> G[MAXN];
int dist[MAXN];
int prevv[MAXN];        //圖的鄰接表
int preve[MAXN];        //最短路的前驅節點和對應的邊

void addedge(int from,int to,int cap,int cost) {
    G[from].pb(edge(to,cap,cost,(int)G[to].size()));
    G[to  ].pb(edge(from,0,-cost,(int)G[from].size()-1));
}

typedef pair<int,int> P;
priority_queue<P,vector<P>, greater<P> > qu;
int h[MAXN];

int min_cost_flow(int s,int t,int f){
    int res = 0;
    fill(h+1,h+1+N,0);
    while(f > 0){
        while (!qu.empty()) qu.pop();
        mmm(dist,0x3f);
        dist[s] = 0; qu.push(P(0,s));
        while(!qu.empty()){
            P now = qu.top(); qu.pop();
            if(dist[now.second] < now.first) continue;
            int v = now.second;
            rep(i,0,G[v].size()-1) {
                edge &e = G[v][i];
                if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
                    dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
                    prevv[e.to] = v;
                    preve[e.to] = i;
                    qu.push(P(dist[e.to],e.to));
                }
            }
        }
        //cout << dist[t] << endl;
        if(dist[t] == INF) return res;
        rep(i,1,N) h[i] += dist[i];
        int d = f;
        for(int v = t; v != s; v = prevv[v])
            d = min(d,G[prevv[v]][preve[v]].cap);
        f -= d;
        res += d * h[t];
        for(int v = t;v != s;v = prevv[v]){
            edge &e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}
void init(){
    rep(i,1,N) G[i].clear();
}

int a,b,PP,Q;
int east[20][20];
int north[20][20];

int main()
{
	scanf("%d %d",&a,&b);
	scanf("%d %d",&PP,&Q);
	N = (PP+1)*(Q+1);
    int s = ++N;
    int t = ++N;

    int num = 1;
	rep(i,1,PP+1) {
        rep(j,1,Q) {
            scanf("%d",&east[i][j]);
            addedge(num,num+1,1,-east[i][j]);
            addedge(num,num+1,INF,0);
            num ++;
        }
        num ++;
	}   
	num = 0;
	rep(i,1,Q+1) {
	    int tmpNum = ++num;
        rep(j,1,PP) {
            scanf("%d",&north[i][j]);
            addedge(tmpNum,tmpNum+Q+1,1,-north[i][j]);
            addedge(tmpNum,tmpNum+Q+1,INF,0);
            tmpNum += Q+1;
        }
	}

	rep(i,1,a) {
        int y,x,k;
        scanf("%d %d %d",&k,&x,&y);
        y ++,x ++;
        int id = (x-1)*(Q+1)+y;
        addedge(s,id,k,0);
	}
	rep(i,1,b) {
        int y,x,k;
        scanf("%d %d %d",&k,&x,&y);
        y ++,x ++;
        int id = (x-1)*(Q+1)+y;
        addedge(id,t,k,0);
	}

    int res = -min_cost_flow(s,t,INF);
    printf("%d\n",res);
}