• 寒假Day16POJ3422Kaka's Matrix Travels(最大费用最大流+拆点)


    思路:最大费用最大流

    AC代码:

      1 #include<string.h>
      2 #include<iostream>
      3 #include<stdio.h>
      4 #include<algorithm>
      5 #include<queue>
      6 #include<vector>
      7 #include<map>
      8 #include<cmath>
      9 using namespace std;
     10 #define inf 0x3f3f3f3f
     11 const int N=55*55*2;
     12 typedef long long ll;
     13 
     14 struct node
     15 {
     16     int to,nextt;
     17     int cap,flow,cost;
     18 } e[N*N];
     19 
     20 bool book[N];
     21 int pre[10*N],head[10*N],dist[N],a[N][N];
     22 int tot,s,t;
     23 
     24 void add(int u,int v,int cap,int cost)
     25 {
     26     e[tot].to=v;
     27     e[tot].cap=cap;
     28     e[tot].cost=cost;
     29     e[tot].flow=0;
     30     e[tot].nextt=head[u];
     31 
     32     head[u]=tot++;
     33     e[tot].to=u;
     34     e[tot].cap=0;
     35     e[tot].cost=-cost;
     36     e[tot].flow=0;
     37     e[tot].nextt=head[v];
     38     head[v]=tot++;
     39 }
     40 
     41 bool SPFA()
     42 {
     43     for(int i=0; i<=t; i++)
     44     {
     45         dist[i]=inf;
     46         book[i]=0;
     47         pre[i]=-1;
     48     }
     49     book[s]=1;
     50     dist[s]=0;
     51     queue<int>Q;
     52     Q.push(s);
     53     while(!Q.empty())
     54     {
     55         int u=Q.front();
     56         Q.pop();
     57         book[u]=0;
     58         for(int i=head[u]; i!=-1; i=e[i].nextt)
     59         {
     60             int v=e[i].to;
     61             if(e[i].cap>e[i].flow&&dist[v]>dist[u]+e[i].cost)
     62             {
     63                 dist[v]=dist[u]+e[i].cost;
     64                 pre[v]=i;
     65                 if(book[v]==0)
     66                 {
     67                     book[v]=1;
     68                     Q.push(v);
     69                 }
     70             }
     71         }
     72     }
     73     if(dist[t]!=inf)
     74         return 1;
     75     return 0;
     76 }
     77 
     78 int MCMF()
     79 {
     80     int flow=0,cost=0;
     81     while(SPFA())
     82     {
     83         int minn=inf;
     84         for(int i=pre[t]; i!=-1; i=pre[e[i^1].to])
     85             minn=min(minn,e[i].cap-e[i].flow);
     86         for(int i=pre[t]; i!=-1; i=pre[e[i^1].to])
     87         {
     88             e[i].flow+=minn;
     89             e[i^1].flow-=minn;
     90             cost+=e[i].cost*minn;
     91         }
     92         flow+=minn;
     93     }
     94     return cost;
     95 }
     96 
     97 int main()
     98 {
     99     int n,k;
    100     while(~scanf("%d %d",&n,&k))
    101     {
    102         for(int i=1;i<=n;i++)
    103         {
    104             for(int j=1;j<=n;j++)
    105             {
    106                 scanf("%d",&a[i][j]);
    107             }
    108         }
    109         tot=0;
    110         memset(head,-1,sizeof(head));
    111         s=0,t=n*n*2+1;//t=n*n+1
    112        /* for(int i=1; i<=n; i++)
    113         {
    114             for(int j=1; j<=n; j++)
    115             {
    116                 if(i==n&&j==n)
    117                     continue;
    118                 if(i==n)//down
    119                     add(i+1,j,1,a[i+1][j]);
    120                 else if(j==n)//right
    121                     add(i,j+1,1,a[i][j+1]);
    122                 else
    123                 {
    124                     add(i,j+1,1,a[i][j+1]);//right
    125                     add(i+1,j,1,a[i+1][j]);//down
    126                 }
    127             }
    128         }*/
    129         for(int i=1;i<=n;i++)
    130         {
    131             for(int j=1;j<=n;j++)
    132             {
    133                 int x=(i-1)*n+j;
    134                 add(x*2,x*2+1,1,-a[i][j]);
    135                 add(x*2,x*2+1,k-1,0);
    136                 if(j!=n)//向下
    137                     add(x*2+1,(x+1)*2,k,0);
    138                 if(i!=n)//
    139                     add(x*2+1,(x+n)*2,k,0);
    140             }
    141         }
    142         add(s,2,k,0);
    143         add(2*n*n+1,t,k,0);
    144         int mincost=MCMF()*(-1);
    145         printf("%d\n",mincost);
    146 
    147     }
    148     return 0;
    149 }
    View Code
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  • 原文地址:https://www.cnblogs.com/OFSHK/p/12247188.html
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