• 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
    " :: "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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yejinru/p/3303281.html
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