POJ 3422 Kaka's Matrix Travels（费用流）

Kaka's Matrix Travels Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6792   Accepted: 2679 Description On an N × N chessboard with a non-negative number in each grid, Kaka starts his matrix travels with SUM = 0. For each travel, Kaka moves one rook from the left-upper grid to the right-bottom one, taking care that the rook moves only to the right or down. Kaka adds the number to SUM in each grid the rook visited, and replaces it with zero. It is not difficult to know the maximum SUM Kaka can obtain for his first travel. Now Kaka is wondering what is the maximum SUM he can obtain after his Kth travel. Note the SUM is accumulative during the K travels. Input The first line contains two integers N and K (1 ≤ N ≤ 50, 0 ≤ K ≤ 10) described above. The following N lines represents the matrix. You can assume the numbers in the matrix are no more than 1000. Output The maximum SUM Kaka can obtain after his Kth travel. Sample Input 3 2 1 2 3 0 2 1 1 4 2 Sample Output 15 Source POJ Monthly--2007.10.06, Huang, Jinsong

 

测试一下最小费用最大流的模板。

 

很经典的建图方法。

 

  1 #include <stdio.h>
  2 #include <algorithm>
  3 #include <string.h>
  4 #include <iostream>
  5 #include <string>
  6 #include <queue>
  7 using namespace std;
  8 
  9 const int MAXN = 10000;
 10 const int MAXM = 100000;
 11 const int INF = 0x3f3f3f3f;
 12 struct Edge
 13 {
 14     int to,next,cap,flow,cost;
 15 }edge[MAXM];
 16 int head[MAXN],tol;
 17 int pre[MAXN],dis[MAXN];
 18 bool vis[MAXN];
 19 int N;//节点总个数，节点编号从0~N-1
 20 void init(int n)
 21 {
 22     N = n;
 23     tol = 0;
 24     memset(head,-1,sizeof(head));
 25 }
 26 void addedge(int u,int v,int cap,int cost)
 27 {
 28     edge[tol].to = v;
 29     edge[tol].cap = cap;
 30     edge[tol].cost = cost;
 31     edge[tol].flow = 0;
 32     edge[tol].next = head[u];
 33     head[u] = tol++;
 34     edge[tol].to = u;
 35     edge[tol].cap = 0;
 36     edge[tol].cost = -cost;
 37     edge[tol].flow = 0;
 38     edge[tol].next = head[v];
 39     head[v] = tol++;
 40 }
 41 bool spfa(int s,int t)
 42 {
 43     queue<int>q;
 44     for(int i = 0;i < N;i++)
 45     {
 46         dis[i] = INF;
 47         vis[i] = false;
 48         pre[i] = -1;
 49     }
 50     dis[s] = 0;
 51     vis[s] = true;
 52     q.push(s);
 53     while(!q.empty())
 54     {
 55         int u = q.front();
 56         q.pop();
 57         vis[u] = false;
 58         for(int i = head[u]; i != -1;i = edge[i].next)
 59         {
 60             int v = edge[i].to;
 61             if(edge[i].cap > edge[i].flow &&
 62                dis[v] > dis[u] + edge[i].cost )
 63             {
 64                 dis[v] = dis[u] + edge[i].cost;
 65                 pre[v] = i;
 66                 if(!vis[v])
 67                 {
 68                     vis[v] = true;
 69                     q.push(v);
 70                 }
 71             }
 72         }
 73     }
 74     if(pre[t] == -1)return false;
 75     else return true;
 76 }
 77 //返回的是最大流，cost存的是最小费用
 78 int minCostMaxflow(int s,int t,int &cost)
 79 {
 80     int flow = 0;
 81     cost = 0;
 82     while(spfa(s,t))
 83     {
 84         int Min = INF;
 85         for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
 86         {
 87             if(Min > edge[i].cap - edge[i].flow)
 88                 Min = edge[i].cap - edge[i].flow;
 89         }
 90         for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
 91         {
 92             edge[i].flow += Min;
 93             edge[i^1].flow -= Min;
 94             cost += edge[i].cost * Min;
 95         }
 96         flow += Min;
 97     }
 98     return flow;
 99 }
100 
101 int a[55][55];
102 int main()
103 {
104     int n,k;
105     while(scanf("%d%d",&n,&k) == 2)
106     {
107         for(int i = 0;i < n;i++)
108             for(int j = 0;j < n;j++)
109                 scanf("%d",&a[i][j]);
110         init(2*n*n+2);
111         for(int i = 0;i < n;i++)
112             for(int j = 0;j < n;j++)
113             {
114                 addedge(n*i+j+1,n*n+n*i+j+1,1,-a[i][j]);
115                 addedge(n*i+j+1,n*n+n*i+j+1,INF,0);
116             }
117 
118         for(int i = 0;i < n;i++)
119             for(int j = 0;j < n;j++)
120             {
121                 if(i < n-1)
122                     addedge(n*n+n*i+j+1,n*(i+1)+j+1,INF,0);
123                 if(j < n-1)
124                     addedge(n*n+n*i+j+1,n*i+j+1+1,INF,0);
125             }
126         addedge(0,1,k,0);
127         addedge(2*n*n,2*n*n+1,INF,0);
128         int cost;
129         minCostMaxflow(0,2*n*n+1,cost);
130         printf("%d\n",-cost);
131     }
132     return 0;
133 }