hdu 1565 方格取数(1)

这个题网上很多人都说用状态压缩dp来做，我就是觉得状态压缩dp有点那么理解不上啊，不过如果这个题吧相邻的两个格子连起来，那不就是求最大权独立点集吗？奋战了三天，我的第一道最大流题目终于写出来了，高兴啊！

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;

const double pi=acos(-1.0);
const double eps=1e-8;
typedef pair<int,int>pii;

const int maxn=2500+10;

struct Edge
{
    int from,to,cap,flow;
};

int n,m,s,t;
vector<Edge>edges;
vector<int>G[maxn];
int d[maxn],cur[maxn];
bool vis[maxn];

void AddEdge(int from,int to,int cap)
{
    Edge temp;
    temp.cap=cap; temp.flow=0; temp.from=from; temp.to=to;
    edges.push_back(temp);
    temp.cap=0; temp.flow=0; temp.from=to; temp.to=from;
    edges.push_back(temp);
    m=edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}

bool BFS()
{
    memset(vis,0,sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s]=0;
    vis[s]=1;
    while(!Q.empty())
    {
        int x=Q.front();Q.pop();
        for (int i=0;i<G[x].size();i++)
        {
            Edge& e=edges[G[x][i]];
            if (!vis[e.to] && e.cap>e.flow)
            {
                vis[e.to]=1;
                d[e.to]=d[x]+1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

int DFS(int x,int a)
{
    if (x==t || a==0) return a;
    int flow=0,f;
    for (int i=cur[x];i<G[x].size();i++)
    {
        Edge& e=edges[G[x][i]];
        if (d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            e.flow+=f;
            edges[G[x][i]^1].flow-=f;
            flow+=f;
            a-=f;
            if (a==0) break;
        }
    }
    return flow;
}

int Dinic()
{
    int flow=0;
    while (BFS())
    {
        memset(cur,0,sizeof(cur));
        flow+=DFS(s,inf);
    }
    return flow;
}

void init()
{
    for (int i=0;i<=maxn;i++) G[i].clear();
    edges.clear();
}

int main()
{
    //freopen("in.txt","r",stdin);

    int N,M,a,sum;
    int dx[4]={-1,1,0,0};
    int dy[4]={0,0,-1,1};
    while (scanf("%d",&N)!=EOF)
    {
        init();
        s=0;t=N*N+1;
        sum=0;
        for (int i=1;i<=N;i++)
        for (int j=1;j<=N;j++)
        {
            scanf("%d",&a);
            sum+=a;
            int x=(i-1)*N+j;
            if ((i+j)%2==0)
            {
                AddEdge(s,x,a);
                for (int k=0;k<4;k++)
                {
                    int xx=i+dx[k];
                    int yy=j+dy[k];
                    if (xx>=1 && xx<=N && yy>=1 && yy<=N)
                    {
                        int y=(xx-1)*N+yy;
                        AddEdge(x,y,inf);
                    }
                }
            }
            else
            {
                AddEdge(x,t,a);
                for (int k=0;k<4;k++)
                {
                    int xx=i+dx[k];
                    int yy=j+dy[k];
                    if (xx>=1 && xx<=N && yy>=1 && yy<=N)
                    {
                        int y=(xx-1)*N+yy;
                        AddEdge(y,x,inf);
                    }
                }
            }
        }
        printf("%d
",sum-Dinic());
    }
    //fclose(stdin);
    return 0;
}