要点,每个点之间可以连两条边,这样可以解决很多问题,该题中的数据被取出,则可以连一条有权重的边和若干条无权重的边.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 10000;
const int MAXM = 1000000;
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 n,k;
int mm[100][100];
int main()
{
// freopen("data.txt","r",stdin);
ios_base::sync_with_stdio(false);
cin >> n>>k;
int s = 0;
int t = n*n*2+1;
init(n*n*2+10);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin >> mm[i][j];
}
}
addedge(s,1,k,0);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
int num =(i-1)*n+j;
addedge( num,num + n*n, 1 ,-mm[i][j] );
addedge( num,num + n*n, INF,0 );
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
int fr =(i-1)*n+j + n*n;
int to =(i)*n+j;
if(i!=n)
addedge(fr,to,INF,0 );
fr =(i-1)*n+j + n*n;
to =(i-1)*n+j+1;
if(j!=n)
addedge(fr,to,INF,0 );
}
}
addedge( 2*n*n,t,INF,0);
int ans =0;
minCostMaxflow(s,t,ans);
cout << -ans << endl;
return 0;
}