洛谷P4012 深海机器人问题费用流

最新推荐文章于 2022-08-09 21:41:48 发布

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最新推荐文章于 2022-08-09 21:41:48 发布

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本文链接：https://blog.youkuaiyun.com/liyong1009s/article/details/87277921

本文详细解析了洛谷P4012深海机器人问题的费用流算法实现，通过构建复杂的图模型，运用SPFA算法求解最大流最小费用路径，解决机器人在海洋地图上的最优路径规划问题。

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洛谷P4012 深海机器人问题费用流

Code:

#include<cstdio>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int maxn=1000;
const int INF=1000000+233;
int mapp[20][20];
int s,t,n,cnt=0;
struct Edge{
	int from,to,cap,cost;
	Edge(int u,int v,int c,int f):from(u),to(v),cap(c),cost(f){}
};
struct MCMF{
    vector<Edge>edges;
    vector<int>G[maxn];
    int d[maxn],inq[maxn],a[maxn],flow2[maxn];
    queue<int>Q;
    ll ans=0;
    int flow=0;
    void addedge(int u,int v,int c,int f){
    	edges.push_back(Edge(u,v,c,f));    //正向弧
    	edges.push_back(Edge(v,u,0,-f));   //反向弧
    	int m=edges.size();
    	G[u].push_back(m-2);
    	G[v].push_back(m-1);
    }
    int SPFA(){
    	for(int i=0;i<=n+4;++i)d[i]=INF,flow2[i]=INF;
        d[t]=INF,flow2[t]=INF;
    	memset(inq,0,sizeof(inq));int f=INF;
    	d[s]=0,inq[s]=1;Q.push(s);
        while(!Q.empty()){
        	int u=Q.front();Q.pop();inq[u]=0;
        	int sz=G[u].size();
        	for(int i=0;i<sz;++i){
                  Edge e=edges[G[u][i]];
                  if(e.cap>0&&d[e.to]>d[u]+e.cost){
                      a[e.to]=G[u][i];
                      d[e.to]=d[u]+e.cost;
                      flow2[e.to]=min(flow2[u],e.cap);
                      if(!inq[e.to]){inq[e.to]=1;Q.push(e.to);}
                  }
        	}
        }
        if(flow2[t]>1||d[t]==INF||d[t]==0)return 0;
        f=flow2[t];
        flow+=f;
        int u=edges[a[t]].from;
        edges[a[t]].cap-=f;
        edges[a[t]^1].cap+=f;
        while(u!=s){
        	edges[a[u]].cap-=f;
        	edges[a[u]^1].cap+=f;
        	u=edges[a[u]].from;
        }
        ans+=(ll)(d[t])*(-1);
        return 1;
    }
    ll maxflow(){
        while(SPFA());
        return ans;
    }
}op;
int main(){
   int p,q,a,b;
   scanf("%d%d",&a,&b);
   scanf("%d%d",&p,&q);
   ++p,++q;
   n=p*q;
   for(int i=0;i<p;++i)
      for(int j=0;j<q;++j)mapp[i][j]=++cnt;
   for(int i=0;i<p;++i)
    for(int j=0;j<q-1;++j){
        int c;scanf("%d",&c);
        op.addedge(mapp[i][j],mapp[i][j+1],1,-c);
        op.addedge(mapp[i][j],mapp[i][j+1],INF,0);
    }
   for(int i=0;i<q;++i)
    for(int j=0;j<p-1;++j){
        int c;scanf("%d",&c);
        op.addedge(mapp[j][i],mapp[j+1][i],1,-c);
        op.addedge(mapp[j][i],mapp[j+1][i],INF,0);
    }
   s=0,t=500;
   for(int i=1;i<=a;++i)
   {
      int k,x,y;
      scanf("%d%d%d",&k,&x,&y);
      op.addedge(s,mapp[x][y],k,0);
   }
   for(int i=1;i<=b;++i){
    int k,x,y;
    scanf("%d%d%d",&k,&x,&y);
    op.addedge(mapp[x][y],t,k,0);
   }
   printf("%lld",op.maxflow());
   return 0;
}