POJ 3422 Kaka's Matrix Travels（费用流）

3 2
1 2 3
0 2 1
1 4 2

15

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <string>
#include <queue>
using namespace std;

const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i = 0;i < N;i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
int flow = 0;
cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
{
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}

int a[55][55];
int main()
{
int n,k;
while(scanf("%d%d",&n,&k) == 2)
{
for(int i = 0;i < n;i++)
for(int j = 0;j < n;j++)
scanf("%d",&a[i][j]);
init(2*n*n+2);
for(int i = 0;i < n;i++)
for(int j = 0;j < n;j++)
{
addedge(n*i+j+1,n*n+n*i+j+1,1,-a[i][j]);
addedge(n*i+j+1,n*n+n*i+j+1,INF,0);
}

for(int i = 0;i < n;i++)
for(int j = 0;j < n;j++)
{
if(i < n-1)
addedge(n*n+n*i+j+1,n*(i+1)+j+1,INF,0);
if(j < n-1)
addedge(n*n+n*i+j+1,n*i+j+1+1,INF,0);
}
addedge(0,1,k,0);
addedge(2*n*n,2*n*n+1,INF,0);
int cost;
minCostMaxflow(0,2*n*n+1,cost);
printf("%d\n",-cost);
}
return 0;
}