HDU 1565 方格取数(1) 状态压缩DP

题目大意：

从n*n的矩阵中取出一些数使得这些数互不相邻，问最大和为多少

大致思路：

明显的状态压缩DP，每两行之间的状态转移，这里受到内存限制只开两个数组来表示当先行和下一行来进行转移，原本想用vector来记录那两个状态之间可以转换的，但是受到内存限制还是用时间换空间了

代码如下：

Result  :  Accepted     Memory  :  8520 KB     Time  :  890 ms

/*
* Author: Gatevin
* Created Time: 2014/8/16 20:52:13
* File Name: 123.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

int dp[1 << 20];
int nex[1 << 20];
vector <int> yes;
//vector <int> G[1 << 20];
int n;
int a[21][21];

bool check(int x)
{
for(int i = 0; i <= (n - 2); i++)
{
if(((x >> i) & 1) == 1 && ((x >> (i + 1)) & 1) == 1)
{
return false;
}
}
return true;
}

void init(int n)
{
for(int i = 0; i <= ( 1 << n) - 1; i++)
{
if(check(i))
{
yes.push_back(i);
}
}
return;
}

int main()
{
while(~scanf("%d", &n))
{
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
scanf("%d", &a[i][j]);
}
}
if(n == 1)
{
printf("%d\n", a[1][1]);
continue;
}
memset(dp, 0, sizeof(dp));
memset(nex, 0, sizeof(nex));
yes.clear();
init(n);
for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
for(int j = 0; j <= n - 1; j++)
{
if((1 << j) & yes[i])
{
dp[yes[i]] += a[1][j + 1];
nex[yes[i]] += a[2][j + 1];
}
}
}
int dpnow = 1;
/*for(int i = 0; i <= (1 << n) - 1; i++)
{
G[i].clear();
}
for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
for(unsigned int j = i + 1; j <= yes.size() - 1; j++)
{
if((yes[i] & yes[j]) == 0)
{
G[yes[i]].push_back(yes[j]);
G[yes[j]].push_back(yes[i]);
}
}
}
*/
while(dpnow < n)
{
/*for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
int tmp = nex[yes[i]];
for(unsigned int j = 0; j <= G[yes[i]].size() - 1; j++)
{
nex[yes[i]] = max(nex[yes[i]], tmp + dp[G[yes[i]][j]]);
}
}
*/
for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
int tmp = nex[yes[i]];
for(unsigned int j = 0; j <= yes.size() - 1; j++)
{
if((yes[i] & yes[j]) == 0)
{
nex[yes[i]] = max(nex[yes[i]], tmp + dp[yes[j]]);
}
}
}
for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
dp[yes[i]] = nex[yes[i]];
}
dpnow++;
if(dpnow == n) break;
memset(nex, 0, sizeof(nex));
for(unsigned int i = 0;i <= yes.size() - 1; i++)
{
for(int j = 0; j <= n - 1; j++)
{
if(yes[i] & (1 << j))
{
nex[yes[i]] += a[dpnow + 1][j + 1];
}
}
}
}
int answer = 0;
for(unsigned int i = 0; i <= yes.size() - 1; i++)
{
answer = max(answer, dp[yes[i]]);
}
printf("%d\n", answer);
}
return 0;
}