http://acm.hdu.edu.cn/showproblem.php?pid=3376
http://acm.hdu.edu.cn/showproblem.php?pid=2686
http://poj.org/problem?id=3422
POJ 3422为从矩阵左上角走到右下角，最多走k次，每个格子里的数字只能算一次，后面可以重复经过，求经过的各个数字的和的最大值。
拆点，入点向出点连流量为1，费用为当前格子负值的边，向下方，右方连边，流量为k，费用为0，起点连流量为1，源点向1连边，m*m向汇点连边，流量为k，费用为0.
跑费用流。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/17 9:42:51
File Name :6.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 100000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=10000;
const int maxm=500100;
struct Edge{
int next,to,cap,cost;
Edge(int _next=0,int _to=0,int _cap=0,int _cost=0){
next=_next;to=_to;cap=_cap;cost=_cost;
}
}edge[maxm];
int head[maxn],vis[maxn],pre[maxn],dis[maxn],n,tol;
void addedge(int u,int v,int cap,int cost){
edge[tol]=Edge(head[u],v,cap,cost);head[u]=tol++;
edge[tol]=Edge(head[v],u,0,-cost);head[v]=tol++;
}
bool spfa(int s,int t){
queue<int> q;
for(int i=0;i<=n;i++)
dis[i]=INF,vis[i]=0,pre[i]=-1;
dis[s]=0;vis[s]=1;q.push(s);
while(!q.empty()){
int u=q.front();q.pop();vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(edge[i].cap&&dis[v]>dis[u]+edge[i].cost){
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v])vis[v]=1,q.push(v);
}
}
}
if(pre[t]==-1)return 0;
return 1;
}
void fun(int s,int t,int &flow,int &cost){
flow=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)MIN=edge[i].cap;
for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
edge[i].cap-=MIN,edge[i^1].cap+=MIN,cost+=edge[i].cost*MIN;
flow+=MIN;
}
}
int a[70][70];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int m,k;
while(~scanf("%d%d",&m,&k)){
for(int i=1;i<=m;i++)
for(int j=1;j<=m;j++)
scanf("%d",&a[i][j]);
memset(head,-1,sizeof(head));tol=0;
for(int i=1;i<=m;i++)
for(int j=1;j<=m;j++){
int tt=(i-1)*m+j;
addedge(tt,tt+m*m,1,-a[i][j]);
addedge(tt,tt+m*m,INF,0);
if(i<m)addedge(tt+m*m,tt+m,k,0);
if(j<m)addedge(tt+m*m,tt+1,k,0);
}
addedge(0,1,k,0);
addedge(2*m*m,2*m*m+1,k,0);
int flow,cost;
n=2*m*m+10;
fun(0,2*m*m+1,flow,cost);
cout<<-cost<<endl;
}
return 0;
}拆点，入点向出点连流量为1，费用为权值的相反数，每一个点向右方，下方连流量为无穷大，费用为0的边，添加1，m*m流量为1的边，表示可以算两次。
跑费用流。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/17 9:42:51
File Name :6.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 100000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=800000;
const int maxm=5001000;
struct Edge{
int next,to,cap,cost;
Edge(int _next=0,int _to=0,int _cap=0,int _cost=0){
next=_next;to=_to;cap=_cap;cost=_cost;
}
}edge[maxm];
int head[maxn],vis[maxn],pre[maxn],dis[maxn],n,tol;
void addedge(int u,int v,int cap,int cost){
edge[tol]=Edge(head[u],v,cap,cost);head[u]=tol++;
edge[tol]=Edge(head[v],u,0,-cost);head[v]=tol++;
}
bool spfa(int s,int t){
queue<int> q;
for(int i=0;i<=n;i++)
dis[i]=INF,vis[i]=0,pre[i]=-1;
dis[s]=0;vis[s]=1;q.push(s);
while(!q.empty()){
int u=q.front();q.pop();vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(edge[i].cap&&dis[v]>dis[u]+edge[i].cost){
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v])vis[v]=1,q.push(v);
}
}
}
if(pre[t]==-1)return 0;
return 1;
}
void fun(int s,int t,int &flow,int &cost){
flow=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)MIN=edge[i].cap;
for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
edge[i].cap-=MIN,edge[i^1].cap+=MIN,cost+=edge[i].cost*MIN;
flow+=MIN;
}
}
int a[700][700];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int m;
while(~scanf("%d",&m)){
for(int i=1;i<=m;i++)
for(int j=1;j<=m;j++)
scanf("%d",&a[i][j]);
memset(head,-1,sizeof(head));tol=0;
for(int i=1;i<=m;i++)
for(int j=1;j<=m;j++){
int tt=(i-1)*m+j;
addedge(tt,tt+m*m,1,-a[i][j]);
if(i<m)addedge(tt+m*m,tt+m,INF,0);
if(j<m)addedge(tt+m*m,tt+1,INF,0);
}
addedge(1,m*m+1,1,0);
addedge(m*m,2*m*m,1,0);
int flow,cost;
n=2*m*m+10;
fun(1,2*m*m,flow,cost);
cout<<-cost<<endl;
}
return 0;
}
POJ 3422 HDU 2686,3376 费用流拆点建图,布布扣,bubuko.com
POJ 3422 HDU 2686,3376 费用流拆点建图
原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/26059895
