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\n",sum-Dinic());

    }

    //fclose(stdin);

    return 0;

}

hdu 1565 方格取数(1),布布扣,bubuko.com