计算机学院大学生程序设计竞赛（2015’12）Pick Game

Pick Game

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 119    Accepted Submission(s): 20

Problem Description

This is a pick game.

On a n*m matrix, each gird has a value. The player could only choose the gird that is adjacent to at least two empty girds (A grid outside the matrix also regard as empty). Adjacent means two girds share a common edge. If one play chooses one gird, he will
get the value and the gird will be empty. They play in turn.



One day, WKC plays this game with ZJS. 

Both of them are clever students, so they will choose the best strategy. 

WKC plays first, and he wants to know the maximal value he could get.

 

Input

There is a positive integer T(1<=T<=50) in the first line, which specifying the number of test cases to follow.

Each test case begins with two numbers n and m ( 2 <= n, m <= 5 ).

Then n lines follow and each lines with m numbers V_ij (0< V_ij <=1000).

 

Output

Output the maximal value WKC could get.

 

Sample Input

1
2 2
9 8
7 6

 

Sample Output

16

 

总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define mod 10007
#define N 10010
#include<vector>
using namespace std;
vector<int> v[N],dp[N];
int n,m,cnt,Map[6][6];
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
bool ok(int x,int y)
{
    if(x<n&&x>=0&&y<m&&y>=0)
        return true;
    return false;
}
int Gets(int st)
{
    int ret=0;
    for(int i=0; i<cnt; i++)
    {
        if(((1<<i)&st)==0)
        {
            ret+=Map[i/m][i%m];
        }
    }
    return ret;
}
int cal(int st)
{
    int ret=0,i;
    for(i=0; i<cnt; i++)
        if((1<<i)&st)
            ret++;
    return ret;
}
int dfs(int st)
{
    int next,ans,i,j,tmp,x,y,xx,yy;
    x=st%mod;
    for(y=v[x].size()-1; y>=0; y--)
    {
        if(v[x][y]==st)
            return dp[x][y];
    }
    if(cal(st)==cnt-1)
        return Gets(st);
    ans=-20006;
    for(i=0; i<cnt; i++)
    {
        if(((1<<i)&st)==0)
        {
            x=i/m;
            y=i%m;
            tmp=0;
            for(j=0; j<4; j++)
            {
                xx=x+dir[j][0];
                yy=y+dir[j][1];
                if(!ok(xx,yy)||(ok(xx,yy)&&(((1<<(xx*m+yy))&st))))
                {
                    tmp++;
                }
            }
            if(tmp>=2)
            {
                next=(1<<i) | st;
                ans=max(ans,Map[x][y]-dfs(next));
            }
        }
    }
    v[st%mod].push_back(st);
    dp[st%mod].push_back(ans);
    return ans;
}
int main()
{
    int t,i,j;
    scanf("%d",&t);
    while(t--)
    {
        for(i=0; i<mod; i++)
        {
            dp[i].clear();
            v[i].clear();
        }
        scanf("%d%d",&n,&m);
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                scanf("%d",&Map[i][j]);
        cnt=n*m;
        printf("%d\n",(dfs(0)+Gets(0))/2);
    }
    return 0;
}

@执念  "@☆但求“❤”安★
下次我们做的一定会更好。。。。



为什么这次的题目是英文的。。。。QAQ...