hdu1565 网络流或状态压缩DP

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define N 1010
#define M 50010
#define inf 1<<30
using namespace std;
struct Edge{
int to,val,next;
}edge[M];
int index
,d
,gap
,e;
void addedge(int from,int to,int val)
{
edge[e].to=to;
edge[e].val=val;
edge[e].next=index[from];
index[from]=e++;
edge[e].to=from;
edge[e].val=0;
edge[e].next=index[to];
index[to]=e++;
}
int source,des,n,m;
//n is the number of point
int dfs(int pos,int flow)
{
if(pos==des)
return flow;
int i,j,v,val,lv,mind,c;
mind=n-1;//初始最小标号为n-1
lv=flow;
for(i=index[pos];i!=-1;i=edge[i].next)
{
v=edge[i].to;
val=edge[i].val;
if(val)
{
if(d[v]+1==d[pos])
{
c=min(lv,val);//对于该点的最小可行流
c=dfs(v,c);
edge[i].val-=c;//更新剩余图
edge[i^1].val+=c;
lv-=c;
if(d[source]>=n)return flow-lv;
if(lv==0) break;
}
if(d[v]<mind)mind=d[v];//找出与pos相连的点的最小标号
}
}
if(lv==flow)//没有找到增广路劲，进行标号更新
{
--gap[d[pos]];
if(!gap[d[pos]])
d[source]=n;
d[pos]=mind+1;
++gap[d[pos]];
}
return flow-lv;
}
int sap(int st,int de)
{
source=st;
des=de;
memset(d,0,sizeof(d));
memset(gap,0,sizeof(gap));
gap[0]=n;//初始标号为0的有n个.
int ans=0;
while(d[st]<n)
{
ans+=dfs(st,inf);
//cout<<d[st]<<endl;
}
return ans;
}
void init()
{
e=0;
memset(index,-1,sizeof(index));
}
int pos(int a,int b)
{
return (a-1)*m+b;
}
int main()
{
int t,i,j;
while(scanf("%d",&m)!=EOF)
{

init();
int w;
int sum=0;
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&w);
sum+=w;
if((i+j)%2==0)
{
addedge(0,pos(i,j),w);
if(i<m)
addedge(pos(i,j),pos(i+1,j),inf);
if(j<m)
addedge(pos(i,j),pos(i,j+1),inf);
if(i>1)
addedge(pos(i,j),pos(i-1,j),inf);
if(j>1)
addedge(pos(i,j),pos(i,j-1),inf);
}
else
addedge(pos(i,j),m*m+1,w);
}
n=m*m+2;
printf("%d\n",sum-sap(0,n-1));
}
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int s[1<<21];
int map[22][22];
int dp[2][1<<21];
int main()
{
int n,i,j,t,k,num=0;;
n=1<<21;
for(i=0;i<=n;i++)//记录可行状态
if((i&(i<<1))==0)
s[num++]=i;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&map[i][j]);
int p=0;
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)//枚举每一行
{
p^=1;//进行滚动数组
for(j=0;j<num;j++)//枚举每行的所有状态
{
if(s[j]>(1<<n))
break;
int sum=0;
for(k=0;k<n;k++) if(s[j]&(1<<k)) sum+=map[i][k];//记录该状态值
for(k=0;k<num;k++)//枚举已经得到的状态转移到该状态能到的最大值
{
if(s[k]>(1<<n))
break;
if(!(s[k]&s[j]))
dp[p][s[j]]=max(dp[p][s[j]],dp[1-p][s[k]]+sum);
}
}
}
int ans=0;
for(i=0;i<num&&s[i]<=(1<<n);i++)//寻找答案
ans=max(ans,dp[p][s[i]]);
printf("%d\n",ans);
}
return 0;
}