POJ 3422 Kaka's Matrix Travels K取方格数

题目：给出n*n的方格矩阵，现在从左上方走m次到右下方，问m次能够获得的最大价值和。

分析：最大费用流。拆点进行限制每个格子只取一次，假设点x拆成 x,xx，右边（假设有）y,yy，下方（假设有）z,zz

　　     点  点    流量 费用

　　则：x , xx , 1 , -a[i][j]

  　　   x , xx , m  , 0

　　　 xx,  y  , m  , 0

　　　 xx , z  , m  , 0 　

用最小费用流增广m次即可

 

#include <set>

#include <map>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <string>

#include <vector>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>



using namespace std;



typedef long long ll;

typedef unsigned long long ull;



#define debug puts("here")

#define rep(i,n) for(int i=0;i<n;i++)

#define rep1(i,n) for(int i=1;i<=n;i++)

#define REP(i,a,b) for(int i=a;i<=b;i++)

#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)

#define pb push_back

#define RD(n) scanf("%d",&n)

#define RD2(x,y) scanf("%d%d",&x,&y)

#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)

#define All(vec) vec.begin(),vec.end()

#define MP make_pair

#define PII pair<int,int>

#define PQ priority_queue

#define cmax(x,y) x = max(x,y)

#define cmin(x,y) x = min(x,y)

#define Clear(x) memset(x,0,sizeof(x))

/*



#pragma comment(linker, "/STACK:1024000000,1024000000")



int size = 256 << 20; // 256MB

char *p = (char*)malloc(size) + size;

__asm__("movl %0, %%esp\n" :: "r"(p) );



*/



/******** program ********************/



char op;

inline void Int(int &x){

    while( !isdigit(op=getchar()) );

    x = op-'0';

    while(isdigit(op=getchar()))

        x = x*10+op-'0';

}



const int MAXN = 5005;

const int MAXM = 200005;

const int INF = 1e9;



int pre[MAXN],dis[MAXN];

int po[MAXN],tol;

bool use[MAXN];

int q[MAXM],head,tail;

int n,m,vs,vt,ans;

int a[55][55];



struct node{

    int y,f,cost,next;

}edge[MAXM];



inline void Add(int x,int y,int f,int cost){

    edge[++tol].y = y;

    edge[tol].f = f;

    edge[tol].cost = cost;

    edge[tol].next = po[x];

    po[x] = tol;

}



inline void add(int x,int y,int f,int cost){

    Add(x,y,f,cost);

    Add(y,x,0,-cost);

}



inline bool spfa(){

    memset(use,false,sizeof(use));

    rep1(i,vt)

        dis[i] = INF;

    dis[vs] = 0;

    head = tail = 0;

    q[tail++] = vs;

    pre[vs] = 0;

    while(head<tail){

        int x = q[head++];

        use[x] = false;

        for(int i=po[x];i;i=edge[i].next){

            int y = edge[i].y;

            if(edge[i].f>0&&edge[i].cost+dis[x]<dis[y]){

                dis[y] = dis[x]+edge[i].cost;

                pre[y] = i;

                if(!use[y]){

                    use[y] = true;

                    q[tail++] = y;

                }

            }

        }

    }

    if(dis[vt]==INF)

        return false;

    int aug = INF;

    for(int i=pre[vt];i;i=pre[edge[i^1].y])

        aug = min(aug,edge[i].f);

    for(int i=pre[vt];i;i=pre[edge[i^1].y]){

        edge[i].f -= aug;

        edge[i^1].f += aug;

    }

    ans += dis[vt]*aug;

    return true;

}



inline int id(int x,int y){

    return (x-1)*n+y;

}



int main(){



#ifndef ONLINE_JUDGE

    freopen("sum.in","r",stdin);

    //freopen("sum.out","w",stdout);

#endif



    while(~RD2(n,m)){

        int nn = n*n;

        vs = 1;

        vt = nn<<1;



        tol = 1;

        Clear(po);



        rep1(i,n)

            rep1(j,n)

                Int(a[i][j]);

                //RD(a[i][j]);



        rep1(i,n){

            rep1(j,n){

                int now = id(i,j);

                add( now,now+nn,1,-a[i][j] );

                add( now,now+nn,m,0 );

                if(i<n) add( now+nn,now+n,m,0 );

                if(j<n) add( now+nn,now+1,m,0 );

            }

        }



        ans = 0;

        while(m--&&spfa())

            ;

        cout<<-ans<<endl;

    }



    return 0;

}