POJ 3422 Kaka's Matrix Travels(费用流)

POJ 3422 Kaka's Matrix Travels(费用流)

http://poj.org/problem?id=3422

题意:

给一个N*N的方阵，从[1,1]到[n,n]走K次，走过每个方格加上上面的数(每个方格初始都有一个非负数)，然后这个格上面的数变为0。求可取得的最大的值。

分析:

其实把每个网格看成一条边的话,那么我们就是等于要找K次最小费用增广路径即可(就是一个求最小费用最大流的过程). 建图如下:

源点s编号0,n*n个网格每个网格分成两个点i和i+n*n, 汇点t编号为n*n*2+1.

从源点s到1号节点有边(s , 1, K)

从每个网格到自己有边(i, i+n*n, 1, -cost) 和(i, i+n*n, INF, 0) (注意这里的-cost,因为原题要我们求最大值)

从每个网格i到它的右或下那个网格j有边(i+n*n, j, INF, 0)

从最右下角到汇点t有边(2*n*n, t, INF, 0)

最终我们求得的最小费用的绝对值就是权值最大值.

AC代码: 最小费用取绝对值解法

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define INF 1e9
using namespace std;
const int maxn= 51*51*2+10;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int from,int to,int cap,int flow,int cost):from(from),to(to),cap(cap),flow(flow),cost(cost){}
};

struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    int d[maxn];
    int a[maxn];
    int p[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;++i) G[i].clear();
    }

    void AddEdge(int from,int to,int cap,int cost)
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int &flow,int &cost)
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;++i) d[i]=INF;
        queue<int> Q;
        d[s]=0,inq[s]=true, Q.push(s),a[s]=INF,p[s]=0;
        while(!Q.empty())
        {
            int u =Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;i<G[u].size();++i)
            {
                Edge &e=edges[G[u][i]];
                if(e.cap>e.flow && d[e.to]>d[u]+e.cost)
                {
                    d[e.to]=d[u]+e.cost;
                    a[e.to]=min(a[u],e.cap-e.flow);
                    p[e.to]=G[u][i];
                    if(!inq[e.to]){inq[e.to]=true; Q.push(e.to);}
                }
            }
        }
        if(d[t]==INF) return false;
        flow += a[t];
        cost += a[t]*d[t];
        int u=t;
        while(u!=s)
        {
            edges[p[u]].flow +=a[t];
            edges[p[u]^1].flow -=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }

    int solve()
    {
        int flow=0,cost=0;
        while(BellmanFord(flow,cost));
        return cost;
    }
}MM;

int mp[55][55];

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)==2)
    {
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
            scanf("%d",&mp[i][j]);

        int src=0,dst=n*n*2+1;
        MM.init(n*n*2+2,src,dst);
        MM.AddEdge(src,1,k,0);
        MM.AddEdge(n*n*2,dst,INF,0);
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
        {
            int id=(i-1)*n+j;
            MM.AddEdge(id,id+n*n,1,-mp[i][j]);
            MM.AddEdge(id,id+n*n,INF,0);

            if(j+1<=n) MM.AddEdge(id+n*n,id+1,INF,0);
            //这里如果写if(id+1<=n*n)是错的,最后一列虽然能+1,但是无后继
            if(i+1<=n) MM.AddEdge(id+n*n,id+n,INF,0);
        }
        printf("%d\n",-MM.solve());
    }
    return 0;
}

AC代码: 直接求最大费用路径解法,两个程序的不同点在下面已经标出来了,注意细节

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define INF 1e9
using namespace std;
const int maxn= 51*51*2+10;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int from,int to,int cap,int flow,int cost):from(from),to(to),cap(cap),flow(flow),cost(cost){}
};

struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    int d[maxn];
    int a[maxn];
    int p[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;++i) G[i].clear();
    }

    void AddEdge(int from,int to,int cap,int cost)
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int &flow,int &cost)
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;++i) d[i]=-1;           //修改1
        queue<int> Q;
        d[s]=0,inq[s]=true, Q.push(s),a[s]=INF,p[s]=0;
        while(!Q.empty())
        {
            int u =Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;i<G[u].size();++i)
            {
                Edge &e=edges[G[u][i]];
                if(e.cap>e.flow && d[e.to]<d[u]+e.cost)//修改2
                {
                    d[e.to]=d[u]+e.cost;
                    a[e.to]=min(a[u],e.cap-e.flow);
                    p[e.to]=G[u][i];
                    if(!inq[e.to]){inq[e.to]=true; Q.push(e.to);}
                }
            }
        }
        if(d[t]==-1) return false;//修改3
        flow += a[t];
        cost += a[t]*d[t];
        int u=t;
        while(u!=s)
        {
            edges[p[u]].flow +=a[t];
            edges[p[u]^1].flow -=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }

    int solve()
    {
        int flow=0,cost=0;
        while(BellmanFord(flow,cost));
        return cost;
    }
}MM;

int mp[55][55];

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)==2)
    {
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
            scanf("%d",&mp[i][j]);

        int src=0,dst=n*n*2+1;
        MM.init(n*n*2+2,src,dst);
        MM.AddEdge(src,1,k,0);
        MM.AddEdge(n*n*2,dst,INF,0);
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
        {
            int id=(i-1)*n+j;
            MM.AddEdge(id,id+n*n,1,mp[i][j]);//修改4
            MM.AddEdge(id,id+n*n,INF,0);

            if(j+1<=n) MM.AddEdge(id+n*n,id+1,INF,0);
            //这里如果写if(id+1<=n*n)是错的,最后一列虽然能+1,但是无后继
            if(i+1<=n) MM.AddEdge(id+n*n,id+n,INF,0);
        }
        printf("%d\n",MM.solve());//修改5
    }
    return 0;
}