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洛谷P2045 方格取數加強版 最小費用流

edge queue spa div urn 取數 for dde inf

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=10000+3;
const int INF=100000+123;
int s,t,n;
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*2;++i)d[i]=INF,flow2[i]=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(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;
    }
};
int main(){
    int siz,k,cnt=0;
    MCMF op;
    scanf("%d%d",&siz,&k);
    n=siz*siz;
    for(int i=1;i<=siz;++i)
        for(int j=1;j<=siz;++j){
             int c;scanf("%d",&c);  
             ++cnt;
             op.addedge(cnt,cnt+1,1,-c);
             op.addedge(cnt,cnt+1,INF,0);
             ++cnt;
        }
    t=cnt;
    cnt=0;
    for(int i=1;i<=siz;++i)
        for(int j=1;j<=siz;++j){
            cnt+=2;
            if(i+1<=siz)op.addedge(cnt,cnt+(siz*2-1),INF,0);
            if(j+1<=siz)op.addedge(cnt,cnt+1,INF,0);
        }
    s=0;
    op.addedge(s,1,k,0);
    printf("%lld",op.maxflow());
    return 0;
}

  

洛谷P2045 方格取數加強版 最小費用流